CHAPTER
10 SIMPLE HARMONIC MOTION AND
ELASTICITY
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1. 0.12 m
2. (c) The restoring force is given by Equation 10.2 as F = −kx, where k is the spring constant
(positive). The graph of this equation is a straight line and indicates that the restoring force
has a direction that is always opposite to the direction of the displacement. Thus, when x is
positive, F is negative, and vice versa.
1 m
. Greater
2π k
values for the mass m and smaller values for the spring constant k lead to greater values for
the period.
3. (b) According to Equations 10.4 and 10.11, the period T is given by T =
4. (d) The maximum speed in simple harmonic motion is given by Equation 10.8 ( vmax = Aω ) .
Thus, increases in both the amplitude A and the angular frequencyω lead to an increase in the
maximum speed.
5. (e) The maximum acceleration in simple harmonic motion is given by Equation 10.10
( amax = Aω 2 ) . A decrease in the amplitude A decreases the maximum acceleration, but this
decrease is more than offset by the increase in the angular frequency ω, which is squared in
Equation 10.10.
6. 1.38 m/s
7. (b) The velocity has a maximum magnitude at point A, where the object passes through the
position where the spring is unstrained. The acceleration at point A is zero, because the
spring is unstrained there and is not applying a force to the object. The velocity is zero at
point B, where the object comes to a momentary halt and reverses the direction of its travel.
The magnitude of the acceleration at point B is a maximum, because the spring is maximally
stretched there and, therefore, applies a force of maximum magnitude to the object.
8. 0.061 m
9. +7.61 m/s2
10. 0.050 m
496 SIMPLE HARMONIC MOTION AND ELASTICITY
11. (c) The principle of conservation of mechanical energy applies in the absence of
nonconservative forces, so that KE + PE = constant. Thus, the total energy is the same at all
points of the oscillation cycle. At the equilibrium position, where the spring is unstrained,
the potential energy is zero, and the kinetic energy is KEmax; thus, the total energy is KEmax.
At the extreme ends of the cycle, where the object comes to a momentary halt, the kinetic
energy is zero, and the potential energy is PEmax; thus, the total energy is also PEmax. Since
both KEmax and PEmax equal the total energy, it must be true that KEmax = PEmax.
12. (e) In simple harmonic motion the speed and, hence, KE has a maximum value as the object
passes through its equilibrium position, which is position 2. EPE has a maximum value
when the spring is maximally stretched at position 3. GPE has a maximum value when the
object is at its highest point above the ground, that is, at position 1.
13. (a) At the instant the top block is removed, the total mechanical energy of the remaining
system is all elastic potential energy and is 12 kA2 (see Equation 10.13), where A is the
amplitude of the previous simple harmonic motion. This total mechanical energy is
conserved, because friction is absent. Therefore, the total mechanical energy of the ensuing
simple harmonic motion is also 12 kA2 , and the amplitude remains the same as it was
k
. Thus, when
m
the mass m attached to the spring decreases, the angular frequency increases.
previously. The angular frequency ω is given by Equation 10.11 as ω =
14. (b) The angular frequency ω of oscillation of a simple pendulum is given by Equation 10.16
⎛
g⎞
⎜⎜ ω =
⎟ . It depends only on the magnitude g of the acceleration due to gravity and the
L ⎟⎠
⎝
length L of the pendulum. It does not depend on the mass. Therefore, the pendulum with the
greatest length has the smallest frequency.
15. 1.7 s
16. (c) When the energy of the system is dissipated, the amplitude of the motion decreases. The
motion is called damped harmonic motion.
17. (a) Resonance occurs when the frequency of the external force equals the frequency of
oscillation of the object on the spring. The angular frequency of such a system is given by
⎛
k ⎞
Equation 10.11 ⎜⎜ ω =
⎟ . Since the frequency of the force is doubled, the new frequency
m ⎟⎠
⎝
must be 2ω = 2
k
8k
k
=2
. The frequency of system A is, in fact, ω =
.
2m
m
m
Chapter 10 Answers to Focus on Concepts Questions
497
⎡
⎛ ΔL ⎞ ⎤
18. (c) According to Equation 10.17 ⎢ F = Y ⎜
A , the force F required to stretch a piece of
⎜ L ⎟⎟ ⎥
⎝ 0 ⎠ ⎦⎥
⎣⎢
material is proportional to Young’s modulus Y, the amount of stretch ∆L, and the crosssectional area A of the material, but is inversely proportional to the initial length L0 of the
FL
material. Solving this equation for the amount of stretch gives Δ L = 0 . Thus, the greater
YA
the cross-sectional area, the smaller is the amount of stretch, for given values of Young’s
modulus, the initial length, and the stretching force. Thus, B stretches more than A, because
B has the smaller cross-sectional area of solid material.
19. 0.50 × 10−6 m
20. 0.0017
498 SIMPLE HARMONIC MOTION AND ELASTICITY
CHAPTER 10 SIMPLE HARMONIC MOTION AND
ELASTICITY
PROBLEMS
______________________________________________________________________________
1.
SSM REASONING AND SOLUTION Using Equation 10.1, we first determine the
spring constant:
F Applied
89.0 N
k= x
=
= 4660 N/m
0.0191 m
x
Again using Equation 10.1, we find that the force needed to compress the spring by
0.0508 m is
FxApplied = kx = (4660 N/m)(0.0508 m) = 237 N
______________________________________________________________________________
2.
REASONING AND SOLUTION According to Equation 10.1 and the data from the graph,
the effective spring constant is
FxApplied
160 N
= 6.7 ×102 N/m
x
0.24 m
______________________________________________________________________________
k=
3.
=
REASONING The force required to stretch the spring is given by Equation 10.1 as
FxApplied = kx, where k is the spring constant and x is the displacement of the stretched spring
from its unstrained length. Solving for the spring constant gives k = FxApplied / x . The force
applied to the spring has a magnitude equal to the weight W of the board, so FxApplied = W .
Since the board’s lower end just extends to the floor, the unstrained length x0 of the spring,
plus the length L0 of the board, plus the displacement x of the stretched spring must equal
the height h0 of the room, or x0 + L0 + x = h0. Thus, x = h0 − x0 − L0.
SOLUTION Substituting FxApplied = W and x = h0 − x0 − L0 into Equation 10.1, we find
k=
FxApplied
x
=
104 N
W
=
= 650 N/m
h0 − x0 − L0 2.44 m − 1.98 m − 0.30 m
____________________________________________________________________________________________
Chapter 10 Problems
4.
499
REASONING The weight of the person causes the spring in the scale to compress. The
amount x of compression, according to Equation 10.1, depends on the magnitude FxApplied of
the applied force and the spring constant k.
SOLUTION
a. Since the applied force is equal to the person’s weight, the spring constant is
Applied
k=
Fx
=
x
670 N
0.79 × 10
4
−2
m
= 8.5 × 10 N / m
(10.1)
b. When another person steps on the scale, it compresses by 0.34 cm. The weight (or
applied force) that this person exerts on the scale is
Applied
Fx
(
4
)(
= k x = 8.5 × 10 N / m 0.34 × 10
−2
)
m = 290 N
(10.1)
______________________________________________________________________________
5.
SSM
REASONING AND SOLUTION According to Newton's second law, the force
required to accelerate the trailer is Fx = max (Equation 4.2a). The displacement x of the
spring is given by Fx = −kx (Equation 10.2). Solving Equation 10.2 for x and using
Fx = max , we obtain
x=−
Fx
k
=−
max
k
=−
( 92 kg ) ( 0.30 m/s2 )
2300 N/m
= −0.012 m
The amount that the spring stretches is 0.012 m .
______________________________________________________________________________
6.
REASONING The restoring force of the spring and the static frictional force point in
opposite directions. Since the box is in equilibrium just before it begins to move, the net
force in the horizontal direction is zero at this instant. This condition, together with the
expression for the restoring force (Equation 10.2) and the expression for the maximum static
frictional force (Equation 4.7), will allow us to determine how far the spring can be
stretched without the box moving upon release.
SOLUTION The drawing at the right shows the four forces
that act on the box: its weight mg, the normal force FN, the
restoring force Fx exerted by the spring, and the maximum
static frictional force fsMAX. Since the box is not moving, it is
in equilibrium. Let the x axis be parallel to the table top.
According to Equation 4.9a, the net force ΣFx in the x direction
must be zero, ΣFx = 0.
+y
FN
fsMAX
Fx
mg
+x
500 SIMPLE HARMONIC MOTION AND ELASTICITY
The restoring force is Fx = –kx (Equation 10.2), where k is the spring constant and x is the
displacement of the spring (assumed to be in the +x direction). The magnitude of the
maximum static frictional force is fsMAX = μsFN (Equation 4.7), where μs is the coefficient
of static friction and FN is the magnitude of the normal force. Thus, the condition for
equilibrium can be written as
μs FN
− k x + μs FN = 0 or x =
k
ΣFx
We can determine FN by noting that the box is also in vertical equilibrium, so that
− mg + FN = 0
FN = mg
or
ΣFy
The distance that the spring is stretched from its unstrained position is
x=
μs FN
=
μs mg
=
( 0.74 )( 0.80 kg ) ( 9.80 m /s 2 )
= 9.8 × 10−2 m
59 N /m
k
k
______________________________________________________________________________
7.
REASONING The block is at equilibrium as it hangs on the spring. Therefore, the
downward-directed weight of the block is balanced by the upward-directed force applied to
the block by the spring. The weight of the block is mg, where m is the mass and g is the
acceleration due to gravity. The force exerted by the spring is given by Equation 10.2
(Fx = −kx), where k is the spring constant and x is the displacement of the spring from its
unstrained length. We will apply this reasoning twice, once to the single block hanging, and
again to the two blocks. Although the spring constant is unknown, we will be able to
eliminate it algebraically from the resulting two equations and determine the mass of the
second block.
SOLUTION Let upward be the positive direction. Setting the weight mg equal to the force
−kx exerted by the spring in each case gives
m1 g = −kx1
and
m1 g + m2 g = −kx2
One hanging block
Two hanging blocks
Dividing the equation on the right by the equation on the left, we can eliminate the unknown
spring constant k and find that
m1 g + m2 g
m1 g
=
−kx2
−kx1
or
1+
m2
m1
=
x2
x1
Chapter 10 Problems
501
It is given that x2/x1 = 3.0. Therefore, solving for the mass of the second block reveals that
⎛x
⎞
m2 = m1 ⎜⎜ 2 − 1⎟⎟ = ( 0.70 kg )( 3.0 − 1) = 1.4 kg
⎝ x1
⎠
______________________________________________________________________________
8.
REASONING The spring with the smaller
spring constant (k1 = 33 N/m) is less stiff
than the other spring (k2 = 59 N/m), and,
therefore, stretches farther (x1 > x2). The end
of the rod attached to the stiffer spring is
higher than the other end by a vertical
distance d = x1 − x2 (see the drawing). The
angle θ that the rod makes with the
horizontal is given by the inverse sine
function:
Unstretched position
x2
x1
L
d
θ
⎛d ⎞
⎝ ⎠
θ = sin −1 ⎜ ⎟
L
(1.4)
where L is the length of the rod.
The amount x that each spring is stretched is given by x = F k (Equation 10.1). In order to
use this relation, we will first need to find the amount of force F exerted on the rod by each
spring. The rod is in equilibrium, so the net torque on it must be zero: Στ = 0
(Equation 9.2). We will choose the rod’s center of gravity, located at the center of the
uniform rod, as the axis of rotation. The rod’s
weight acts at this point, and so generates no
Rotation axis
F2
torque. This leaves only the torques due to the two
spring forces, which must sum to zero. Both
L/2
torques have the same lever arm about the center F1
1
θ
of gravity: 1 = 2 = 12 L cos θ (see the drawing at
θ
right) Both forces, therefore, must also have equal
magnitudes, which can be seen as follows:
Στ = F2
( 12 L cos θ ) − F1 ( 12 L cos θ ) = 0
L/2
or
2
F2 = F1
Together, the two vertical spring forces support the rod’s weight. Because their magnitudes
are equal, each force must support half the rod’s weight: F1 = F2 = 12 mg , where m is the
mass of the rod, and g is the magnitude of the acceleration due to gravity.
502 SIMPLE HARMONIC MOTION AND ELASTICITY
SOLUTION As noted above, each spring stretches by an amount given by x = F k
(Equation 10.1). Therefore, the difference d in the heights between the high and low ends of
the rod is
F F
d = x1 − x2 = 1 − 2
k1 k2
Now, using the fact that both force magnitudes are equal to half the weight of the rod
F1 = F2 = 12 mg , we obtain
(
)
d=
1 mg
2
k1
−
1 mg
2
k2
=
mg ⎛ 1 1 ⎞
⎜ − ⎟
2 ⎜⎝ k1 k2 ⎟⎠
Thus, using Equation 1.4 for the angle that the rod makes with the horizontal, we find
⎡ mg ⎛ 1
⎛d ⎞
1 ⎞⎤
θ = sin −1 ⎜ ⎟ = sin −1 ⎢ ⎜⎜ − ⎟⎟ ⎥
⎝L⎠
⎢⎣ 2 L ⎝ k1 k2 ⎠ ⎥⎦
= sin
9.
SSM
(
)
⎡ (1.4 kg ) 9.80 m/s 2
⎤
⎛ 1
1 ⎞⎥
−
⎜
⎟ = 7.0
⎢
2 ( 0.75 m )
⎝ 33 N/m 59 N/m ⎠ ⎥
⎣
⎦
−1 ⎢
REASONING AND SOLUTION The force that acts on the block is given by
Newton's Second law, Fx = max (Equation 4.2a). Since the block has a constant
acceleration, the acceleration is given by Equation 2.8 with v0 = 0 m/s; that is, ax = 2d/t2,
where d is the distance through which the block is pulled. Therefore, the force that acts on
the block is given by
2md
Fx = ma x = 2
t
The force acting on the block is the restoring force of the spring. Thus, according to
Equation 10.2, Fx = − kx , where k is the spring constant and x is the displacement. Solving
Equation 10.2 for x and using the expression above for Fx, we obtain
x=–
Fx
2md
2(7.00 kg)(4.00 m)
=– 2 =–
= − 0.240 m
k
kt
(415 N/m)(0.750 s) 2
The amount that the spring stretches is 0.240 m .
______________________________________________________________________________
Chapter 10 Problems
503
10. REASONING The free-body diagram shows the
+y
Fsmax = μsFN
magnitudes and directions of the forces acting on the
+x
block. The weight mg acts downward. The
max
maximum force of static friction fs acts upward
FN
just before the block begins to slip. The force from F Applied = kx
x
the spring FxApplied = kx (Equation 10.1) is directed
mg
to the right. The normal force FN from the wall
points to the left. The magnitude of the maximum
force of static friction is related to the magnitude of the normal force according to
Equation 4.7 ( fsmax = μs FN ) , where μs is the coefficient of static friction. Since the block is
at equilibrium just before it begins to slip, the forces in the x direction must balance and the
forces in the y direction must balance. The balance of forces in the two directions will
provide two equations, from which we will determine the coefficient of static friction.
SOLUTION Since the forces in the x direction and in the y direction must balance, we have
FN = kx
and
mg = μs FN
Substituting the first equation into the second equation gives
mg = μs FN = μs ( kx )
or
2
mg (1.6 kg ) ( 9.80 m/s )
μs =
=
= 0.79
kx ( 510 N/m )( 0.039 m )
11. SSM REASONING When the ball is whirled in a horizontal circle of radius r at speed v,
the centripetal force is provided by the restoring force of the spring. From Equation 5.3, the
2
magnitude of the centripetal force is mv / r , while the magnitude of the restoring force is
kx (see Equation 10.2). Thus,
mv 2
(1)
= kx
r
The radius of the circle is equal to ( L0 + ΔL ), where L0 is the unstretched length of the
spring and ΔL is the amount that the spring stretches. Equation (1) becomes
mv 2
= k ΔL
L0 + ΔL
(1')
504 SIMPLE HARMONIC MOTION AND ELASTICITY
If the spring were attached to the ceiling and the ball were allowed to hang straight down,
motionless, the net force must be zero: mg − kx = 0 , where –kx is the restoring force of the
spring. If we let Δy be the displacement of the spring in the vertical direction, then
mg = kΔy
Solving for Δy, we obtain
Δy =
mg
k
(2)
SOLUTION According to equation (1') above, the spring constant k is given by
mv 2
ΔL( L0 + ΔL)
k=
Substituting this expression for k into equation (2) gives
Δy =
or
Δy =
mgΔL(L0 + ΔL)
mv
2
=
gΔL(L0 + ΔL)
v2
(9.80 m/s2 )(0.010 m)(0.200 m + 0.010 m)
2
= 2.29 × 10 –3 m
(3.00 m/s)
______________________________________________________________________________
12. REASONING AND SOLUTION The figure at the right
shows the original situation before the spring is cut. The
weight W of the object stretches the string by an amount x.
Applying FxApplied = kx (Equation 10.1) to this situation, (in
which Fx
Applied
kx
= W), gives
W = kx
(1)
W
The figure at the right shows the situation after the spring is
cut into two segments of equal length.
Let k' represent the spring constant of each half of the
spring after it is cut. Now the weight W of the object
stretches each segment by an amount x'.
Applying W = kx to this situation gives
W = k'x' + k'x' = 2k'x'
Combining Equations (1) and (2) yields
k'x'
k'x'
W
(2)
Chapter 10 Problems
505
kx = 2k'x'
From Conceptual Example 2 we know that k' = 2k, so that
kx = 2(2k)x'
Solving for x' gives
x 0.160 m
=
= 0.040 m
4
4
______________________________________________________________________________
x' =
13. REASONING AND SOLUTION From the drawing given with the problem statement, we
see that the kinetic frictional force on the bottom block (#1) is given by
fk1 = µk(m1 + m2)g
(1)
and the maximum static frictional force on the top block (#2) is
fs2MAX = μs m2 g
(2)
Newton’s second law applied to the bottom block gives
F – fk1 – kx = 0
(3)
Newton’s second law applied to the top block gives
fs2MAX − kx = 0
(4)
a. To find the compression x, we have from Equation (4) that
x = fs2MAX /k = µsm2g/k = (0.900)(15.0 kg)(9.80 m/s2)/(325 N/m) = 0.407 m
b. Solving Equation (3) for F and then using Equation (1) to substitute for fk1, we find that
F = kx + fk1 = kx + µk(m1 + m2)g
F = (325 N/m)(0.407 m) + (0.600)(45.0 kg)(9.80 m/s2) = 397 N
______________________________________________________________________________
14. REASONING AND SOLUTION In phase 1 of the block's motion (uniform acceleration)
we find that the net force on the block is F1 − fk = ma where the force of friction is fk = µkmg.
Therefore F1 = m(a + µkg), which is just the force exerted by the spring on the block, i.e.,
F1 = kx1. So we have
506 SIMPLE HARMONIC MOTION AND ELASTICITY
kx1 = m(a + µkg)
(1)
We can find the acceleration using
a =
vf − v0 5.00 m/s − 0 m/s
= 10.0 m/s 2
=
t
0.500 s
In phase 2 of the block's motion (constant speed) a = 0 m/s2, so the force exerted by the
spring is
k x2 = mμk g
(2)
so that
μk =
k x2
mg
Using this expression for μk in Equation (1) we obtain
⎡
⎛ kx ⎞ ⎤
k x1 = m ⎢ a + ⎜ 2 ⎟ g ⎥
⎝ mg ⎠ ⎦
⎣
a. Solving for k gives
k=
(15.0 kg ) (10.0 m/s 2 ) = 1.00 × 103 N/m
ma
=
x1 − x2 0.200 m − 0.0500 m
b. Substituting this value for k into Equation (2) , we have
3
k x2 (1.00 × 10 N/m ) ( 0.0500 m )
=
= 0.340
mg
(15.0 kg ) ( 9.80 m/s 2 )
______________________________________________________________________________
μk =
15. SSM
According to Equations 10.6 and 10.11,
REASONING AND SOLUTION
2 π f = k / m . According to the data in the problem, the frequency of vibration of either
spring is
5.0 cycles 5.0
f =
=
Hz
3.0 s
3.0
Squaring both sides of the equation 2 π f =
2
⎞
k = 4π f m = 4π ⎜
Hz ⎟ ( 320 kg ) = 3.5 × 104 N/m
⎝ 3.0
⎠
______________________________________________________________________________
2
2
2 ⎛ 5.0
k / m and solving for k, we obtain
Chapter 10 Problems
507
16. REASONING The frequency f of the eardrum’s vibration is related to its angular frequency
ω via ω = 2π f (Equation 10.6). The maximum speed during vibration is given by
vmax = Aω (Equation 10.8). We will find the frequency f in part a from Equations 10.6 and
10.8. In part b, we will find the maximum acceleration that the eardrum undergoes with the
aid of amax = Aω 2 (Equation 10.10).
SOLUTION
a. Substituting ω = 2π f (Equation 10.6) into vmax = Aω (Equation 10.8) and solving for
the frequency f, we obtain
vmax = Aω = A ( 2π f )
f =
or
vmax
2π A
=
2.9 × 10−3 m/s
(
2π 6.3 × 10−7 m
)
= 730 Hz
b. Substituting ω = 2π f (Equation 10.6) into amax = Aω 2 (Equation 10.10) yields the
maximum acceleration of the vibrating eardrum:
(
)
2
amax = Aω 2 = A ( 2π f ) = 6.3 ×10−7 m ⎡⎣ 2π ( 730 Hz ) ⎤⎦ = 13 m/s2
2
17. REASONING The force Fx that the spring exerts on the block just before it is released is
equal to –kx, according to Equation 10.2. Here k is the spring constant and x is the
displacement of the spring from its equilibrium position. Once the block has been released,
it oscillates back and forth with an angular frequency given by Equation 10.11 as
ω = k / m , where m is the mass of the block. The maximum speed that the block attains
during the oscillatory motion is vmax = Aω (Equation 10.8). The magnitude of the maximum
acceleration that the block attains is amax = Aω2 (Equation 10.10).
SOLUTION
a. The force Fx exerted on the block by the spring is
Fx = − kx = − ( 82.0 N /m )( 0.120 m ) = −9.84 N
(10.2)
b. The angular frequency ω of the resulting oscillatory motion is
ω=
82.0 N /m
k
=
= 10.5 rad /s
m
0.750 kg
(10.11)
508 SIMPLE HARMONIC MOTION AND ELASTICITY
c. The maximum speed vmax is the product of the amplitude and the angular frequency:
vmax = Aω = ( 0.120 m )(10.5 rad /s ) = 1.26 m /s
(10.8)
d. The magnitude amax of the maximum acceleration is
amax = Aω = ( 0.120 m )(10.5 rad /s ) = 13.2 m /s
2
2
2
(10.10)
______________________________________________________________________________
18. REASONING AND SOLUTION
a. Since the object oscillates between ± 0.080 m , the motion’s amplitude of the motion is
0.080 m .
b. From the graph, the period is T = 4.0 s . Therefore, according to Equation 10.4,
ω=
2π
2π
=
= 1.6 rad/s
T
4.0 s
c. Equation 10.11 relates the angular frequency to the spring constant: ω = k / m . Solving
for k we find
k = ω 2 m = (1.6 rad/s) 2 (0.80 kg) = 2.0 N/m
d. At t = 1.0 s , the graph shows that the spring has its maximum displacement. At this
location, the object is momentarily at rest, so that its speed is v = 0 m/s .
e. The acceleration of the object at t = 1.0 s is a maximum, and its magnitude is
a max = A ω 2 = (0.080 m)(1.6 rad/s) 2 = 0.20 m/s 2
______________________________________________________________________________
19. REASONING The amplitude of simple harmonic motion is the distance from the
equilibrium position to the point of maximum height. The angular frequency ω is related to
the period T of the motion by Equation 10.6. The maximum speed attained by the person is
the product of the amplitude and the angular speed (Equation 10.8).
SOLUTION
a. Since the distance from the equilibrium position to the point of maximum height is the
amplitude A of the motion, we have that A = 45.0 cm = 0.450 m .
b. The angular frequency is inversely proportional to the period of the motion:
Chapter 10 Problems
ω=
2π
2π
=
= 3.31 rad /s
1.90 s
T
509
(10.6)
c. The maximum speed vmax attained by the person on the trampoline depends on the
amplitude A and the angular frequency ω of the motion:
vmax = Aω = ( 0.450 m )( 3.31 rad /s ) = 1.49 m /s
(10.8)
______________________________________________________________________________
20. REASONING The spring constant k of either spring is related to the mass m of the object
attached to it and the angular frequency ω of its oscillation by ω = k m (Equation 10.11).
Squaring both sides of Equation 10.11 and solving for the mass m, we find that m = k ω 2 .
The masses m of the two objects are unknown but identical, so we eliminate m and obtain
m=
k1
ω12
=
k2
ω22
k2 =
or
k1ω22
ω12
(1)
To deal with the angular frequencies, we turn to the maximum velocities. The magnitude
vmax of the maximum velocity of either object is given by vmax = Aω (Equation 10.8),
where A is the amplitude of the object’s motion. Both objects have a maximum velocity of
the same magnitude, so we see that
vmax = A1ω1 = A2ω2
or
ω2 =
A1ω1
A2
(2)
SOLUTION The amplitude of the motion of the mass attached to spring 1 is twice that of
the motion of the mass attached to spring 2: A1 = 2A2. With this substitution, Equation (2)
becomes
2 A2 ω1
ω2 =
= 2ω1
(3)
A2
Substituting Equation (3) into Equation (1) yields
k2 =
k1ω22
ω12
=
(
k1 2 ω1
ω12
)
2
= 4k1 = 4 (174 N/m ) = 696 N/m
510 SIMPLE HARMONIC MOTION AND ELASTICITY
21. SSM REASONING The frequency of vibration of the spring is related to the added
mass m by Equations 10.6 and 10.11:
f =
1
2π
k
m
(1)
The spring constant can be determined from Equation 10.1.
SOLUTION Since the spring stretches by 0.018 m when a 2.8-kg object is suspended from
its end, the spring constant is, according to Equation 10.1,
k=
FxApplied
mg (2.8 kg)(9.80 m/s 2 )
=
=
= 1.52 × 103 N/m
x
x
0.018 m
Solving Equation (1) for m, we find that the mass required to make the spring vibrate at
3.0 Hz is
k
1.52 × 103 N/m
m=
=
= 4.3 kg
4π 2 (3.0 Hz)2
4π 2 f 2
______________________________________________________________________________
22. REASONING The object’s maximum speed vmax occurs when it passes through the
position where the spring is unstrained. The next instant when its maximum acceleration
amax occurs is when it stops momentarily at either x = +A or x = −A, where A is the
amplitude of the motion. The time t that elapses between these two instants is one fourth of
the period T, so the time we seek is
t = 14 T
(1)
The period is given by T = 2π ω (Equation 10.4), where ω is the angular frequency of the
oscillations. To determine the angular frequency, we note that both the maximum speed vmax
and the maximum acceleration amax depend upon the angular frequency: vmax = Aω
(Equation 10.8) and amax = Aω 2 (Equation 10.10). Substituting Equation 10.8 into
Equation 10.10, we eliminate the amplitude A and obtain a direct relationship between the
maximum speed vmax, the maximum acceleration amax, and the angular frequency ω:
amax = Aω 2 = ( Aω ) ω = vmaxω
SOLUTION Substituting Equation (2) into T =
2π
ω
or
ω=
amax
vmax
(Equation 10.4) yields
(2)
Chapter 10 Problems
T=
2π
⎛ amax
⎜⎜
⎝ vmax
⎞
⎟⎟
⎠
2π vmax
=
511
(3)
amax
Then, substituting Equation (3) into Equation (1), we obtain the time t between maximum
speed and maximum acceleration:
⎛ 2π vmax
t = 14 ⎜
⎜ a
⎝ max
⎞ π vmax π (1.25 m/s )
=
= 0.285 s
⎟⎟ =
2
⎠ 2amax 2 6.89 m/s
(
)
23. REASONING Since air resistance is being ignored and the bungee cord is assumed to be
an ideal spring, the bungee jumper oscillates up and down in simple harmonic motion. The
angular frequency ω of oscillation is related to the spring constant k and the mass m of the
jumper by Equation 10.11: ω = k / m . Solving for the spring constant gives
k = mω 2
(1)
The angular frequency ω is inversely proportional to the period T of oscillation according to
ω = 2π / T (see Equation 10.4). Substituting this expression for ω into Equation (1) yields
⎛ 2π ⎞
k = mω = m ⎜
⎟
⎝ T ⎠
2
2
The mass of the jumper is known. The period of oscillation can be obtained from the fact
that the jumper makes two complete oscillations in a time of 9.6 s.
SOLUTION Since the bungee jumper moves through 2 complete cycles in 9.6 s, the time to
complete one cycle (the period of oscillation) is T = 12 ( 9.6 s ) = 4.8 s . The spring constant of
the bungee cord is
2
2
⎛ 2π ⎞
⎛ 2π ⎞
k = m⎜
⎟ = ( 82 kg ) ⎜
⎟ = 140 N/m
⎝ T ⎠
⎝ 4.8 s ⎠
______________________________________________________________________________
24. REASONING AND SOLUTION
The cup slips when the force of static friction is
overcome. So F = ma = µsmg, where the acceleration is the maximum value for the simple
harmonic motion, i.e., so that
amax = Aω2 = A(2π f )2
(10.10)
μs = A(2π f )2/g = (0.0500 m)4π2(2.00 Hz)2/(9.80 m/s2) = 0.806
______________________________________________________________________________
512 SIMPLE HARMONIC MOTION AND ELASTICITY
25. REASONING The work done in stretching or compressing a spring is given directly by
(
2
2
)
Equation 10.12 as W = 12 k x0 − xf , where k is the spring constant and x0 and xf are,
respectively, the initial and final displacements of the spring from its equilibrium position.
The work is positive if the restoring force and the displacement have the same direction and
negative if they have opposite directions.
SOLUTION
a. The work done in stretching the spring from +1.00 to +3.00 m is
(
)
W = 12 k x02 − xf2 =
1
2
( 46.0 N /m ) ⎡⎢⎣(1.00 m )2 − ( 3.00 m )2 ⎤⎥⎦ =
−1.84 × 102 J
b. The work done in stretching the spring from –3.00 m to +1.00 m is
(
)
W = 12 k x02 − xf2 =
1
2
( 46.0 N /m ) ⎡⎢⎣( −3.00 m )2 − (1.00 m )2 ⎤⎥⎦ =
+1.84 × 102 J
c. The work done in stretching the spring from –3.00 to +3.00 m is
(
)
W = 12 k x02 − xf2 =
1
2
( 46.0 N /m ) ⎡⎢⎣( −3.00 m )2 − ( 3.00 m )2 ⎤⎥⎦ =
0J
______________________________________________________________________________
26. REASONING We will find the work done by the spring force from Welastic = 12 kx02 − 12 kxf2
(Equation 10.12), where k is the spring constant and x0 and xf are the initial and final
compressions of the spring, measured relative to its unstrained length. The distances given
in the problem are measured in millimeters, and must be converted to meters before using in
Equation 10.12 with k = 250 N/m.
SOLUTION The spring is initially compressed 5.0 mm, so we have
⎛ 1m
x0 = 5.0 mm ⎜
⎝ 1000 mm
(
)
⎞
−3
⎟ = 5.0 ×10 m
⎠
To extrude the tip of the pen requires an additional compression of 6.0 mm, so the final
compression of the spring is
⎛ 1m
xf = x0 + 6.0 mm = 5.0 mm + 6.0 mm = 11.0 mm ⎜
⎝ 1000 mm
(
)
⎞
−3
⎟ = 11.0 ×10 m
⎠
Chapter 10 Problems
513
Therefore, from Equation 10.12, the work done by the spring force is
(
Welastic = 12 kx02 − 12 kxf2 = 12 k x02 − xf2
=
1
2
)
( 250 N/m ) ⎡⎢( 5.0 ×10−3 m )
⎣
2
(
)
2⎤
− 11.0 × 10−3 m ⎥ = −0.012 J
⎦
27. REASONING As the block falls, only two forces act on it: its weight and the elastic force
of the spring. Both of these forces are conservative forces, so the falling block obeys the
principle of conservation of mechanical energy. We will use this conservation principle to
determine the spring constant of the spring. Once the spring constant is known, Equation
10.11, ω = k / m , may be used to find the angular frequency of the block’s vibrations.
SOLUTION
a. The conservation of mechanical energy states that the final total mechanical energy Ef is
equal to the initial total mechanical energy E0, or Ef = E0 (Equation 6.9a). The expression
for the total mechanical energy of an object oscillating on a spring is given by Equation
10.14. Thus, the conservation of total mechanical energy can be written as
1
2
m vf2 + 12 I ωf2 + m g hf + 12 k yf2 =
1
2
m v02 + 12 I ω02 + m g h0 + 12 k y02
Ef
E0
Before going any further, let’s simplify this equation by noting which variables are zero.
Since the block starts and ends at rest, vf = v0 = 0 m/s. The block does not rotate, so its
angular speed is zero, ωf = ω0 = 0 rad/s. Initially, the spring is unstretched, so that y0 = 0 m.
Setting these terms equal to zero in the equation above gives
m g hf + 12 k yf2 = m g h0
Solving this equation for the spring constant k, we have that
k=
mg ( h0 − hf )
1 y2
2 f
=
( 0.450 kg ) ( 9.80 m /s2 ) ( 0.150 m )
1
2
( 0.150 m )2
= 58.8 N /m
b. The angular frequency ω of the block’s vibrations depends on the spring constant k and
the mass m of the block:
58.8 N /m
k
=
= 11.4 rad /s
(10.11)
m
0.450 kg
______________________________________________________________________________
ω=
514 SIMPLE HARMONIC MOTION AND ELASTICITY
28. REASONING The elastic potential energy of an ideal spring is given by Equation 10.13
PE elastic = 12 ky 2 , where k is the spring constant and y is the amount of stretch or
compression from the spring’s unstrained length. Thus, we need to determine x for an object
of mass m hanging stationary from the spring. Since the object is stationary, it has no
acceleration and is at equilibrium. The upward pull of the spring balances the downwardacting weight of the object. We can use this fact to determine y.
SOLUTION The magnitude of the pull of the spring is given by ky, according to Equation
10.2. The weight of the object is mg. Since these two forces must balance, we have
ky = mg or y = mg/k. Substituting this result into Equation 10.13 for the elastic potential
energy gives
2
m2 g 2
⎛ mg ⎞
PE elastic = 12 ky 2 = 12 k ⎜
=
⎟
2k
⎝ k ⎠
Applying this expression to the two spring/object systems, we find
m12 g 2
PE1 =
2k
and
m22 g 2
PE 2 =
2k
Here, we have omitted the subscript “elastic” for convenience. Dividing the right-hand
equation by the left-hand equation gives
m22 g 2
PE 2
m2
= 22k 2 = 22
PE1 m1 g
m1
2k
⎛ m2 ⎞
( 5.0 kg ) = 4.4 J
PE 2 = PE1 ⎜⎜ 22 ⎟⎟ = (1.8 J )
2
( 3.2 kg )
⎝ m1 ⎠
2
or
29. SSM REASONING AND SOLUTION If we neglect air resistance, only the conservative
forces of the spring and gravity act on the object. Therefore, the principle of conservation
of mechanical energy applies.
When the 2.00 kg object is hung on the end of the vertical spring, it stretches the spring by
an amount y, where
y=
F mg (2.00 kg)(9.80 m/s 2 )
=
=
= 0.392 m
k
k
50.0 N/m
(10.1)
This position represents the equilibrium position of the system with the 2.00-kg object
suspended from the spring. The object is then pulled down another 0.200 m and released
from rest ( v0 = 0 m/s). At this point the spring is stretched by an amount of
Chapter 10 Problems
515
0.392 m + 0.200 m = 0.592 m . This point represents the zero reference level ( h = 0 m) for
the gravitational potential energy.
h = 0 m: The kinetic energy, the gravitational potential energy, and the elastic potential
energy at the point of release are:
KE = 12 mv02 = 12 m(0 m/s) 2 = 0 J
PEgravity = mgh = mg (0 m) = 0 J
PE elastic = 12 ky02 = 12 (50.0 N/m)(0.592 m)2 = 8.76 J
The total mechanical energy E0 at the point of release is the sum of the three energies
above: E0 = 8.76 J .
h = 0.200 m: When the object has risen a distance of h = 0.200 m above the release point,
the spring is stretched by an amount of 0.592 m – 0.200 m = 0.392 m . Since the total
mechanical energy is conserved, its value at this point is still E = 8.76 J . The
gravitational and elastic potential energies are:
PE gravity = mgh = (2.00 kg)(9.80 m/s 2 )(0.200 m) = 3.92 J
PE elastic = 12 ky 2 = 12 (50.0 N/m)(0.392 m) 2 = 3.84 J
Since KE + PE gravity + PE elastic = E ,
KE = E – PE gravity – PEelastic = 8.76 J – 3.92 J – 3.84 J = 1.00 J
h = 0.400 m: When the object has risen a distance of h = 0.400 m above the release point,
the spring is stretched by an amount of 0.592 m – 0.400 m = 0.192 m . At this point, the total
mechanical energy is still E = 8.76 J . The gravitational and elastic potential energies
are:
PE gravity = mgh = (2.00 kg)(9.80 m/s 2 )(0.400 m) = 7.84 J
PE elastic = 12 ky 2 = 12 (50.0 N/m)(0.192 m) 2 = 0.92 J
The kinetic energy is
516 SIMPLE HARMONIC MOTION AND ELASTICITY
KE = E – PE gravity – PE elastic = 8.76 J – 7.84 J – 0.92 J = 0 J
The results are summarized in the table below:
h
KE
PEgrav
PEelastic
E
0m
0.200 m
0.400 m
0J
1.00 J
0.00 J
0J
3.92 J
7.84 J
8.76 J
3.84 J
0.92 J
8.76 J
8.76 J
8.76 J
______________________________________________________________________________
30. REASONING Since air resistance is negligible, we can apply the principle of conservation
of mechanical energy, which indicates that the total mechanical energy of the block and the
spring is the same at the instant it comes to a momentary halt on the spring and at the instant
the block is dropped. Gravitational potential energy is one part of the total mechanical
energy, and Equation 6.5 indicates that it is mgh for a block of mass m located at a height h
relative to an arbitrary zero level. This dependence on h will allow us to determine the
height at which the block was dropped.
SOLUTION The conservation of mechanical energy states that the final total mechanical
energy Ef is equal to the initial total mechanical energy E0. The expression for the total
mechanical energy for an object on a spring is given by Equation 10.14, so that we have
1
2
mvf2 + 12 I ωf2 + mghf + 12 kyf2 = 12 mv02 + 12 I ω02 + mgh0 + 12 ky02
Ef
E0
The block does not rotate, so the angular speeds ωf and ω0 are zero. Since the block comes
to a momentary halt on the spring and is dropped from rest, the translational speeds vf and v0
are also zero. Because the spring is initially unstrained, the initial displacement y0 of the
spring is likewise zero. Thus, the above expression can be simplified as follows:
mghf + 12 kyf2 = mgh0
The block was dropped at a height of h0 − hf above the compressed spring. Solving the
simplified energy-conservation expression for this quantity gives
kyf2
( 450 N/m )( 0.025 m ) = 0.048 m or 4.8 cm
h0 − hf =
=
2mg 2 ( 0.30 kg ) ( 9.80 m/s 2 )
2
Chapter 10 Problems
517
31. REASONING Assuming that friction and air resistance are negligible, we can apply the
principle of conservation of mechanical energy, which indicates that the total mechanical
energy of the rod (or ram) and the spring is the same at the instant it contacts the staple and
at the instant the spring is released. Kinetic energy 12 mv 2 is one part of the total mechanical
energy and depends on the mass m and the speed v of the rod. The dependence on the speed
will allow us to determine the speed of the ram at the instant of contact with the staple.
SOLUTION The conservation of mechanical energy states that the final total mechanical
energy Ef is equal to the initial total mechanical energy E0. The expression for the total
mechanical energy for a spring/mass system is given by Equation 10.14, so that we have
1
2
mvf2 + 12 I ωf2 + mghf + 12 kyf2 = 12 mv02 + 12 I ω02 + mgh0 + 12 ky02
Ef
E0
Since the ram does not rotate, the angular speeds ωf and ω0 are zero. Since the ram is
initially at rest, the initial translational speed v0 is also zero. Thus, the above expression can
be simplified as follows:
1
2
mvf2 + mghf + 12 kyf2 = mgh0 + 12 ky02
In falling from its initial height of h0 to its final height of hf, the ram falls through a distance
of h0 − hf = 0.022 m, since the spring is compressed 0.030 m from its unstrained length to
begin with and is still compressed 0.008 m when the ram makes contact with the staple.
Solving the simplified energy-conservation expression for the final speed vf gives
vf =
k ( y02 − yf2 )
m
+ 2 g ( h0 − hf )
( 32 000 N/m ) ⎡⎣( 0.030 m ) − ( 0.008 m )
2
=
0.140 kg
2
⎤
⎦ + 2 9.80 m/s 2 0.022 m = 14 m/s
)
(
)(
32. REASONING Since air resistance is negligible, we can apply the principle of conservation
of mechanical energy, which indicates that the total mechanical energy of the pellet/spring
system is the same when the pellet comes to a momentary halt at the top of its trajectory as
it is when the pellet is resting on the compressed spring. The fact that the total mechanical
energy is conserved will allow us to determine the spring constant.
SOLUTION The conservation of mechanical energy states that the final total mechanical
energy Ef is equal to the initial total mechanical energy E0. The expression for the total
mechanical energy for an object on a spring is given by Equation 10.14, so that we have
518 SIMPLE HARMONIC MOTION AND ELASTICITY
1 mv 2
f
2
+ 12 I ωf2 + mghf + 12 kyf2 =
1 mv 2
0
2
+ 12 I ω02 + mgh0 + 12 ky02
E
(1)
E
f
0
The pellet does not rotate, so the angular speeds ωf and ω0 are zero. Since the pellet is at rest
as it sits on the spring and since the pellet comes to a momentary halt at the top of its
trajectory, the translational speeds v0 and vf are also zero. Because the spring is unstrained
when the pellet reaches its maximum height, the final displacement yf of the spring is
likewise zero. Thus, Equation (1) simplifies to :
mghf = mgh0 + 12 ky02
Solving this simplified energy-conservation expression for the spring constant k and noting
that the pellet rises to distance of hf − hi = 6.10 m above its position on the compressed
spring, we find that
k=
2mg ( hf − h0 )
y02
=
2 ( 2.10 ×10−2 kg ) ( 9.80 m/s 2 ) ( 6.10 m )
( 9.10 ×10−2 m )2
= 303 N/m
33. SSM REASONING The only force that acts on the block along the line of motion is the
force due to the spring. Since the force due to the spring is a conservative force, the
principle of conservation of mechanical energy applies. Initially, when the spring is
2
unstrained, all of the mechanical energy is kinetic energy, (1/ 2)mv0 . When the spring is
fully compressed, all of the mechanical energy is in the form of elastic potential energy,
2
(1/ 2)k x max
, where xmax , the maximum compression of the spring, is the amplitude A.
Therefore, the statement of energy conservation can be written as
1
2
2
1
mv0 = 2 kA
2
This expression may be solved for the amplitude A.
SOLUTION Solving for the amplitude A, we obtain
mv02
(1.00 × 10–2 kg)(8.00 m/s)2
=
= 7.18 × 10 –2 m
k
124 N/m
______________________________________________________________________________
A=
Chapter 10 Problems
519
34. REASONING It is assumed in Example 16 that the only forces acting on the jumper are the
gravitational force (his weight) and, for the latter part of his descent, the elastic force of the
bungee cord. Therefore, only conservative forces are present, and we may use energy
conservation to guide our solution. He possesses gravitational potential energy with respect
to the water and elastic potential energy. At the lowest point in his fall, he has no kinetic
energy since he comes to a momentary halt and has zero speed.
SOLUTION The conservation of mechanical energy states that the final total mechanical
energy Ef is equal to the initial total mechanical energy E0, or Ef = E0 (Equation 6.9a). The
expression for the total mechanical energy of an object is given by Equation 10.14. Thus,
the conservation of total mechanical energy can be written as
1
2
m vf2 + 12 I ωf2 + m g hf + 12 k yf2 =
1
2
m v02 + 12 I ω02 + m g h0 + 12 k y02
Ef
E0
We can simplify this equation by noting which variables are zero. The jumper starts from
rest and momentarily comes to a halt at the bottom of the jump; thus, v0 = vf = 0 m/s. He
does not rotate, so his angular speed is zero; ωf = ω0 = 0 rad/s. Initially, the bungee cord is
unstretched, so that y0 = 0 m. Setting these terms to zero in the equation above gives
2
mghf + 12 k yf = mgh0
At his lowest point, the bungee cord is stretched in the downward direction, which is taken
to be the negative direction. Thus, the stretch yf is negative, and we note from Figure 10.36
that yf = hf − hA, where hA is 37.0 m. Substituting this expression for yf into the equation
above and rearranging terms, we find that
2mgh0
⎛ 2mg
⎞
hf2 + ⎜
− 2hA ⎟ hf + hA2 −
=0
k
⎝ k
⎠
c
b
where b =
and
(
2 ( 68.0 kg ) 9.80 m /s
66.0 N/m
c = ( 37.0 m ) −
Thus, we have that
2
2
) − 2 (37.0 m ) = − 53.8 m
(
2 ( 68.0 kg ) 9.80 m /s
66.0 N/m
2
) ( 46.0 m ) = 440.1 m
hf2 − ( 53.8 m ) hf + ( 440.1 m 2 ) = 0
This is a quadratic equation in the variable hf, and its solution is
2
520 SIMPLE HARMONIC MOTION AND ELASTICITY
hf =
− ( −53.8 m ) ±
( −53.8 m )2 − 4 ( 440.1 m 2 )
2
= 45.2 m or 10.1 m
The 45.2-m answer is discarded, because it implies that the jumper comes to a halt at a
distance of only 46.0 m – 45.2 m = 0.81 m below the platform, which is above the point
where the bungee cord is stretched. Thus, when he reaches the lowest point in his fall, his
height above the water is 10.1 m .
______________________________________________________________________________
35. REASONING Since the surface is frictionless, we can apply the principle of conservation
of mechanical energy, which indicates that the final total mechanical energy Ef of the object
and spring is equal to their initial total mechanical energy E0 ; Ef = E0. This conservation
equation, and the fact that the angular frequency ω of the oscillation is related to the spring
constant k and mass m by ω = k / m (Equation 10.11), will permit us to find the speed of
the object at the instant when the spring is stretched by 0.048 m.
SOLUTION Equating the total mechanical energy of the system at the instant the spring is
stretched by xf = +0.048 m to the total mechanical energy when the spring is compressed by
x0 = −0.065 m, we have
1 mv 2
f
2
+ 12 I ωf2 + mghf + 12 kxf2 =
1
2
mv02 + 12 I ω02 + mgh0 + 12 kx02
E
(1)
E
f
0
The object does not rotate, so the angular speeds ωf and ω0 are zero. Since the object is
initially at rest, v0 = 0 m/s. Finally, we note that the height of the object does not change
during the motion, so hf = h0. Thus, Equation (1) simplifies to
1 mv 2
f
2
+ 12 kxf2 = 12 kx02
Solving this expression for the final speed gives
vf =
k
x 2 − xf2
m 0
We now recognize that the term k / m is the angular frequency ω of the motion (see
Equation 10.11). With this substitution, the final speed becomes
vf =
k
2
2
x02 − xf2 = ω x02 − xf2 = (11.3 rad/s ) ( −0.065 m ) − ( 0.048 m ) = 0.50 m/s
m
_____________________________________________________________________________________________
Chapter 10 Problems
521
36. REASONING
a. As the block rests stationary in its equilibrium position, it has no acceleration. According
to Newton’s second law, the net force acting on the block is, therefore, zero. This means
that the downward-directed weight of the block must be balanced by an upward-directed
force. This upward force is the restoring force of the spring and is produced because the
spring is compressed. The compression must be enough for the spring to exert on the block
a restoring force that has a magnitude equal to the block’s weight. This balancing of forces
will allow us to determine the magnitude of the spring’s compression.
b. As the block falls downward after being released, its speed is changing in the manner
characteristic of simple harmonic motion. The block is not in equilibrium, and the forces
acting on it do not balance to zero. Instead of thinking about forces, we may think about
mechanical energy and its conservation. When the block is released from rest, the energy of
the spring/block system is all in the form of gravitational potential energy. Being at rest, the
block has no initial kinetic energy. It also has no initial elastic potential energy, since the
spring is unstrained initially. When the block comes to a momentary halt at the lowest point
in its fall, the energy is all in the form of elastic potential energy. Since the block is again at
rest, it again has no kinetic energy. The spring has been compressed, and gravitational
potential energy has been converted entirely into elastic potential energy. The amount by
which the spring is compressed is determined by the amount of gravitational potential
energy that must be converted into elastic potential energy. The amount must be enough that
the elastic potential energy equals the gravitational potential energy. Thus, we will use
energy conservation to determine the magnitude of the spring’s compression.
The compression of the spring is greater in the non-equilibrium case than in the equilibrium
case. The reason is that in the non-equilibrium case, the block has been allowed to move,
and its inertia carries it beyond its stationary equilibrium position on the spring. The
compression of the spring must increase beyond that corresponding to the stationary
equilibrium position in order to produce the force that is needed to decelerate the block to a
momentary halt.
SOLUTION
a. As the block rests stationary on the spring, the downward-directed weight balances the
upward-directed restoring force from the spring. The magnitude of the weight is mg, and
the magnitude of the restoring force is given by Equation 10.2 without the minus sign as kx.
Thus, we have
2
mg ( 0.64 kg ) ( 9.80 m/s )
mg
=
kx
or x =
=
= 0.037 m
k
170 N/m
Magnitude of
the weight
Magnitude of
the spring force
b. The conservation of mechanical energy states that the final total mechanical energy Ef is
equal to the initial total mechanical energy E0. The expression for the total mechanical
energy for an object on a spring is given by Equation 10.14, so that we have
522 SIMPLE HARMONIC MOTION AND ELASTICITY
1
2
mvf2 + 12 Iω f2 + mghf + 12 kxf2 = 12 mv02 + 12 Iω 02 + mgh0 + 12 kx02
Ef
E0
The block does not rotate, so the angular speeds ωf and ω0 are zero. Since the block comes
to a momentary halt on the spring and is released from rest, the translational speeds vf and v0
are also zero. Because the spring is initially unstrained, the initial displacement x0 of the
spring is likewise zero. Thus, the above expression can be simplified as follows:
mghf + 12 kxf2 = mgh0
or
1
2
kxf2 = mg ( h0 − hf )
The term h0 − hf is the amount by which the spring has compressed, or h0 − hf = xf. Making
this substitution into the simplified energy-conservation equation gives
1
2
kxf2 = mg ( h0 − hf ) = mgxf
Solving for xf, we find
x=
or
1
2
kxf = mg
2
2mg 2 ( 0.64 kg ) ( 9.80 m/s )
=
= 0.074 m
k
170 N/m
As expected, the spring compresses more in the non-equilibrium situation.
k
m
(Equation 10.11), where k is the spring constant and m is the mass of the object. However,
we are given neither k nor m. Instead, we are given information about how much the spring
is compressed and the launch speed of the object. Once launched, the object has kinetic
energy, which is related to its speed. Before launching, the spring/object system has elastic
potential energy, which is related to the amount by which the spring is compressed. This
suggests that we apply the principle of conservation of mechanical energy in order to use the
given information. This principle indicates that the total mechanical energy of the system is
the same after the object is launched as it is before the launch. The resulting equation will
provide us with the value of k/m that we need in order to determine the angular frequency
k
from ω =
.
m
37. SSM
REASONING
The angular frequency ω (in rad/s) is given by ω =
SOLUTION The conservation of mechanical energy states that the final total mechanical
energy Ef is equal to the initial total mechanical energy E0. The expression for the total
mechanical energy for a spring/mass system is given by Equation 10.14, so that we have
Chapter 10 Problems
1
2
523
mvf2 + 12 Iω f2 + mghf + 12 kxf2 = 12 mv02 + 12 Iω 02 + mgh0 + 12 kx02
Ef
E0
Since the object does not rotate, the angular speeds ωf and ω0 are zero. Since the object is
initially at rest, the initial translational speed v0 is also zero. Moreover, the motion takes
place horizontally, so that the final height hf is the same as the initial height h0. Lastly, the
spring is unstrained after the launch, so that xf is zero. Thus, the above expression can be
simplified as follows:
2
k vf
2
2
1
1
mv
=
kx
or
=
2
2
f
0
m x02
Substituting this result into Equation 10.11 shows that
vf2 vf 1.50 m/s
k
ω=
= 2 = =
= 24.2 rad/s
m
x0 x0 0.0620 m
38. REASONING As the sphere oscillates vertically, it is subject to conservative forces only:
gravity and the elastic spring force. Therefore, its total mechanical energy
E = 12 mv 2 + 12 I ω 2 + mgh + 12 ky 2 (Equation 10.14) is conserved, and we use the variable y to
denote the vertical stretching of the spring relative to its unstrained length. We will apply
the energy conservation principle to find the spring constant k.
SOLUTION In terms of the final and initial total mechanical energies Ef and E0, the
conservation principle gives us the following starting point:
1 mv 2
f
2
+ 12 I ωf2 + mghf + 12 kyf2 = 12 mv02 + 12 I ω02 + mgh0 + 12 ky02
E
(1)
E0
f
We note that the sphere does not rotate, so the angular speeds ω0 and ωf are zero.
Equation (1) then becomes
1 mv 2
f
2
+ mghf + 12 kyf2 = 12 mv02 + mgh0 + 12 ky02
(2)
Solving Equation (2) for the spring constant k, we obtain
1k
2
(
yf2
−
y02
)= (
1m
2
v02
− vf2
) + mg ( h0 − hf )
or
k=
(
)
m v02 − vf2 + 2mg ( h0 − hf )
yf2 − y02
(3)
524 SIMPLE HARMONIC MOTION AND ELASTICITY
As the spring stretches from y0 = 0.12 m to yf = 0.23 m, the sphere moves downward by
yf − y0 = 0.11 m, so the difference between the sphere’s initial and final heights is
h0 − hf = 0.11 m, which is positive since h0 is greater than hf. Therefore, from Equation (3),
the spring constant k is
k=
( 0.60 kg ) ⎡⎣( 5.70 m/s )2 − ( 4.80 m/s )2 ⎤⎦ + 2 ( 0.60 kg ) ( 9.80 m/s2 ) (0.11 m)
( 0.23 m )2 − ( 0.12 m )2
= 180 N/m
39. REASONING Since the
x=0m
surface is frictionless, we
v0
can apply the principle of
conservation of mechanical
energy, which indicates
that the total mechanical
0.050 m
energy of the spring/mass
system is the same at the
0.080 m
instant the block contacts the bottle (the final state of the system) and at the instant shown in
the drawing (the initial state). Kinetic energy 12 mv 2 is one part of the total mechanical
energy, and depends on the mass m and the speed v of the block. The dependence of the
kinetic energy on speed is critical to our solution. In order for the block to knock over the
bottle, it must at least reach the bottle. When launched with the minimum speed v0 shown in
the drawing, the block will reach the bottle with a final speed of vf = 0 m/s. We will obtain
the desired initial speed v0 by solving the energy-conservation equation for this variable.
SOLUTION The conservation of mechanical energy states that the final total mechanical
energy Ef is equal to the initial total mechanical energy E0. The expression for the total
mechanical energy for a spring/mass system is given by Equation 10.14, so that we have
1
2
mvf2 + 12 Iω f2 + mghf + 12 kxf2 = 12 mv02 + 12 Iω 02 + mgh0 + 12 kx02
Ef
E0
Since the block does not rotate, the angular speeds ωf and ω0 are zero. Moreover, the block
reaches the bottle with a final speed of vf = 0 m/s when the block is launched with the
minimum initial speed v0. In addition, the surface is horizontal, so that the final and initial
heights, hf and h0, are the same. Thus, the above expression can be simplified as follows:
1
2
kxf2 = 12 kx02 + 12 mv02
In this result, we are given no values for the spring constant k and the mass m. However, we
are given a value for the angular frequency ω. This frequency is given by Equation 10.11
Chapter 10 Problems
525
⎛
k ⎞
⎜⎜ ω =
⎟⎟ , which involves only the ratio k/m. Therefore, in solving the simplified energym
⎝
⎠
conservation expression for the speed v0, we will divide both sides by m, so that the ratio
k/m can be expressed using Equation 10.11.
1
2
kxf2
m
Substituting ω =
v0 = ω
=
1
2
kx02 + 12 mv02
or
m
⎛k⎞
v0 = ⎜ ⎟ ( xf2 − x02 )
⎝m⎠
k
from Equation 10.11, we find
m
(x
2
f
− x02 ) = ( 7.0 rad/s )
( 0.080 m ) − ( 0.050 m )
2
2
= 0.44 m/s
40. REASONING AND SOLUTION Using
f =
1
1
=
= 4.00 Hz
T 0.250 s
and also
f =
1
2π
k
m
we can find the ratio
k
= 4π 2 f 2 = 632 N/(kg ⋅ m)
m
With the object resting on the spring, Fy = ky = mg so that,
g
= 0.0155 m
k
m
When the mass leaves the spring, potential energy of the spring has been converted to
gravitational energy, i.e.,
1 ky ′2 = mgh
2
y=
where
y' = 0.0500 m + 0.0155 m = 0.0655 m
Solving for h we get
⎡ (0.0655 m) 2 ⎤
⎛ k ⎞ ⎛ y'2 ⎞
= 0.138 m
h = ⎜ ⎟ ⎜⎜
⎟⎟ = [ 632 N/(kg ⋅ m) ] ⎢
2 ⎥
⎝ m ⎠ ⎝ 2g ⎠
⎣ 2(9.80 m/s ) ⎦
______________________________________________________________________________
526 SIMPLE HARMONIC MOTION AND ELASTICITY
41. SSM REASONING Using the principle of conservation of mechanical energy, the initial
elastic potential energy stored in the elastic bands must be equal to the sum of the kinetic
energy and the gravitational potential energy of the performer at the point of ejection:
1
k x2
2
1
2
= mv02 + mgh
where v0 is the speed of the performer at
the point of ejection and, from the figure at
the right, h = x sin θ.
x
h
θ
Thus,
1
k x2
2
1
2
= mv02 + mgxsin θ
(1)
From the horizontal motion of the performer
where
v0x = v0 cosθ
(2)
s
v0 x = x
t
(3)
and sx = 26.8 m. Combining equations (2) and (3) gives
sx
v0 =
t(cos θ )
Equation (1) becomes:
s2x
1
1
2
k
x
=
m
+ mgxsin θ
2
2 t 2 cos 2 θ
This expression can be solved for k, the spring constant of the firing mechanism.
SOLUTION Solving for k yields:
2
⎛ sx ⎞
2 mg(sin θ )
k = m⎜
⎟ +
⎝ x t cosθ ⎠
x
⎡
⎤
26.8 m
k = (70.0 kg) ⎢
⎥
⎣ (3.00 m)(2.14 s)( cos 40.0°) ⎦
(
2
)
2 ( 70.0 kg ) 9.80 m/s 2 ( sin 40.0° )
= 2.37 × 103 N/m
3.00 m
______________________________________________________________________________
+
Chapter 10 Problems
527
42. REASONING AND SOLUTION Use conservation of energy to find the speed of point A
(take the pivot to have zero gravitational PE).
Eup = mgh = Edown =
1 Iω2
2
where the moment of inertia of the bar is I =
1
3
+
1 ky2
2
mL2, L = bar length, and ω = v/L.
Substituting these into the energy equation, noting from the drawing accompanying the
problem statement that y =
h=
v=
1 L,
2
( 0.100 m )2 + ( 0.200 m )2 − ( 0.100 m ) = 0.124 m
and that
and solving for v, we find that
(
3 mgL − ky 2
m
)=
(
)
2
3 ⎡⎢( 0.750 kg ) 9.80 m/s 2 ( 0.200 m ) − ( 25.0 N/m )( 0.124 m ) ⎤⎥
⎣
⎦ = 2.08 m/s
0.750 kg
______________________________________________________________________________
43. REASONING As the ball swings down, it reaches it greatest speed at the lowest point in
the motion. One complete cycle of the pendulum has four parts: the downward motion in
which the ball attains its greatest speed at the lowest point, the subsequent upward motion in
which the ball slows down and then momentarily comes to rest. The ball then retraces its
motion, finally ending up where it originally began. The time it takes to reach the lowest
point is one-quarter of the period of the pendulum, or t = (1/4)T. The period is related to the
angular frequency ω of the pendulum by Equation 10.4, T = 2π/ω. Thus, the time for the
ball to reach its lowest point is
t = 14 T =
1 ⎛ 2π ⎞
⎜
⎟
4⎝ ω ⎠
The angular frequency ω of the pendulum depends on its length L and the acceleration g due
to gravity through the relation ω = g / L (Equation 10.16). Thus, the time is
1 ⎛ 2π
t= ⎜
4⎝ ω
⎞ 1 ⎛ 2π ⎞ π
⎟= ⎜
⎟= 2
⎠ 4
g
⎜⎜
⎟⎟
⎝ L⎠
L
g
SOLUTION After the ball is released, the time that has elapsed before it attains its greatest
speed is
π L
π
0.65 m
t=
=
= 0.40 s
2 g
2 9.80 m/s 2
______________________________________________________________________________
528 SIMPLE HARMONIC MOTION AND ELASTICITY
44. REASONING The magnitude g of the acceleration due to gravity is related to the length L
and frequency f of the simple pendulum by 2π f = g L (Equation 10.16). Squaring both
sides of Equation 10.16 and solving for g, we obtain
4π 2 f 2 =
g
L
g = 4π 2 f 2 L
or
(1)
The frequency f is the number of complete vibrations made per second. In measuring the
frequency of the simple pendulum, the astronauts recorded N complete vibrations occurring
over a total elapsed time t. The pendulum’s oscillation frequency is, then, given by
f =
N
t
(2)
SOLUTION Substituting Equation (2) into Equation (1), we obtain
2
N⎞
4π 2 N 2 L 4π 2 (100 ) (1.2 m )
g = 4π ⎜ ⎟ L =
=
= 6.0 m/s2
2
2
t
⎝ t ⎠
( 280 s )
2
2⎛
45. REASONING
a. The angular frequency ω of a simple pendulum can be found directly from Equation
10.16 as ω = g / L , where g is the magnitude of the acceleration due to gravity and L is the
length of the pendulum.
b. The total mechanical energy of the pendulum as it swings back and forth is the
gravitational potential energy it has just before it is released, since the pendulum is released
from rest and has no initial kinetic energy. The reason is that friction is being neglected, and
the tension in the cable is always perpendicular to the motion of the bob, so the tension does
no work. Thus, the work done by nonconservative forces, such as friction and tension, is
zero. This means that the total mechanical energy is conserved (see Equation 6.9b) and is
the same at all points along the motion, including the initial point where the bob is released.
c. To find the speed of the bob as it passes through the lowest point of the swing, we will
use the conservation of energy, which relates the total mechanical energy at the lowest point
to that at the highest point.
SOLUTION
a. The angular frequency of the pendulum is
ω=
g
9.80 m/s 2
=
= 3.5 rad/s
L
0.79 m
(10.16)
Chapter 10 Problems
b. At the moment the pendulum is released, the only type of energy it
has is its gravitational potential energy. Thus, its potential energy PE
is equal to its initial total mechanical energy E0, so PE = E0.
According to Equation 6.5, the potential energy of the pendulum is
PE = mgh, where m is the mass of the bob and h is its height above its
equilibrium position (i.e., its position when the pendulum hangs
straight down). The drawing shows that this height is related to the
length L of the pendulum by h = L (1 − cos8.50° ) . Thus, the total
mechanical energy of the pendulum is
529
8.50°
L
h
E0 = mgh = mgL (1 − cos8.50° )
= ( 0.24 kg ) ( 9.80 m/s 2 ) ( 0.79 m )(1 − cos8.50° ) = 2.0 × 10−2 J
c. As the bob passes through the lowest point of the swing, it has only kinetic energy, so its
total mechanical energy is Ef = 12 mvf2 . Since the total mechanical energy is conserved
( Ef
= E0 ) , we have that
1
2
mvf2 = E0
Solving for the final speed gives
2 ( 2.0 × 10−2 J )
= 0.41 m/s
m
0.24 kg
______________________________________________________________________________
vf =
2 E0
=
46. REASONING The length L of a simple pendulum is related to its frequency f via
g
2π f =
(Equation 10.16). In terms of its period T, the frequency of a simple pendulum
L
is f =
2π
1
=
(Equation 10.5), so we have
T
T
T
=
2π
L
g
or
g
. Solving for the length L, we obtain
L
T2
L
=
2
g
4π
or
L =
T 2g
4π 2
(1)
We will use Equation (1) to calculate the difference L2 − L1 between the final and initial
lengths of the pendulum.
530 SIMPLE HARMONIC MOTION AND ELASTICITY
SOLUTION The period of the original pendulum is T1 = 1.25 s. When its length is
increased from L1 to L2, its period increases by 0.20 s to T2 = 1.25 s + 0.20 s = 1.45 s. From
Equation (1), the difference L2 − L1 between the pendulum’s final and initial lengths is
L2 − L1 =
T22 g
4π 2
−
T12 g
4π 2
=
g
4π
(
T2
2 2
− T12
)
9.80 m/s 2 ⎡
2
2
=
1.45 s ) − (1.25 s ) ⎤ = 0.13 m
(
2
⎣
⎦
4π
47. REASONING According to Equation 10.15, the angular frequency ω of a physical
pendulum is ω = mgL / I and depends on the ratio of the mass m to the moment of inertia
I. Since the moment of inertia is directly proportional to the mass (see Equation 9.6), the
mass algebraically cancels. Thus, the angular frequency is independent of the mass of the
physical pendulum. According to Equation 10.4, the period is T = 2π /ω. Since the angular
frequency ω is independent of the mass, so is the period. These two expressions will allow
us to determine the periods of the wood and metal pendulums.
SOLUTION
a. The period T of a pendulum is given by Equation 10.4 as T = 2π/ω, where ω is its
angular frequency. The angular frequency of a physical pendulum is given by Equation
10.15 as ω = mgL / I , where m is its mass, L is the distance from the pivot to the center of
mass, and I is the moment of inertia about the pivot. Combining these two relations yields
T=
2π
ω
=
2π
mgL
I
I
mgL
= 2π
The moment of inertia of a meter stick (a thin rod) that is oscillating about an axis at one
end is given in Table 9.1 as I = 13 mL20 , where L0 is the length of the stick. Since the meter
stick is uniform, the distance L from one end to its center of mass is L = 12 L0 . Therefore, the
period of oscillation of the wood pendulum is
2
1
mL0
I
3
= 2π
= 2π
T = 2π
mgL
mg 12 L0
(
= 2π
2 (1.00 m )
(
3 9.80 m /s
2
)
)
2 L0
3g
= 1.64 s
b. The period is the same for the metal pendulum, since the mass has been eliminated
algebraically in the expression for T.
______________________________________________________________________________
Chapter 10 Problems
531
48. REASONING For small-angle displacements, the frequency of simple harmonic motion for
a physical pendulum is determined by 2 π f = mgL / I , where L is the distance between the
axis of rotation and the center of gravity of the rigid body of moment of inertia I. Since the
frequency f and the period T are related by f = 1/ T , the period of pendulum A is given by
TA = 2π
I
mgL
Since the pendulum is made from a thin, rigid, uniform rod, its moment of inertia is given
by I = (1/ 3)md 2 , where d is the length of the rod. Since the rod is uniform, its center of
gravity lies at its geometric center, and L = d / 2 Therefore, the period of pendulum A is
given by
2d
TA = 2π
3g
For the simple pendulum we have
d
TB = 2π
g
SOLUTION The ratio of the periods is, therefore,
TA 2π 2d /(3g)
2
=
=
= 0.816
3
TB
2π d / g
______________________________________________________________________________
49. REASONING The relation between the period T and angular frequency ω is T =
2π
ω
mgL
I
(Equation 10.15), where m is the mass of the pendulum, g is the acceleration due to gravity,
L is the distance between the axis of rotation at the pivot point and the center of gravity of
the rod, and I is the moment of inertia of the rod. According to Table 9.1, the moment of
inertia of a thin uniform rod of length D is I = 13 mD 2 . Combining these three equations
algebraically will give us an expression for the period that we seek. However, the length D
of the rod is not given. Instead, the period of the simple pendulum is given. We will be
able to use this information to eliminate the need for the missing length data.
(Equation 10.6). The angular frequency of a physical pendulum is given by ω =
SOLUTION Substituting Equation 10.15 for ω into Equation 10.6, shows that the period of
the physical pendulum is
2π
2π
I
T=
=
= 2π
mgL
ω
mgL
I
532 SIMPLE HARMONIC MOTION AND ELASTICITY
Now we can use the expression I = 13 mD 2 for the moment of inertia of the rod. In addition,
we recognize that the center of gravity of the uniform rod lies at the center of the rod, so that
L = 12 D . With these two substitutions the expression for the period becomes
2
1
I
2D
3 mD
= 2π
= 2π
T = 2π
1
mgL
mg ( 2 D )
3g
(1)
At this point, we must deal with the unknown length D of the rod. To this end, we note that
the period of the simple pendulum is given by Equations 10.6 and 10.16 as
ωSimple =
2π
TSimple
g
D
=
or
TSimple = 2π
D
g
Solving this expression for D/g and substituting the result into Equation (1) gives
⎛ 2⎞
⎛ 2⎞
T = ⎜⎜
⎟⎟ TSimple = ⎜⎜
⎟⎟ ( 0.66 s ) = 0.54 s
⎝ 3⎠
⎝ 3⎠
50. REASONING AND SOLUTION The period of the sphere (a physical pendulum) is
Ts = 2π
where for a solid sphere I =
7 MR2
5
I
mgL
(see Table 9.1) and L = R. Therefore,
Ts = 2π
7R
5g
For the simple pendulum we know that
To = 2π
L
g
We want Ts = To which leads to
7R L
or
L = 75 R
=
5g g
______________________________________________________________________________
Chapter 10 Problems
533
51. SSM REASONING When the tow truck pulls the car out of the ditch, the cable stretches
and a tension exists in it. This tension is the force that acts on the car. The amount ΔL that
the cable stretches depends on the tension F, the length L0 and cross-sectional area A of the
cable, as well as Young’s modulus Y for steel. All of these quantities are given in the
statement of the problem, except for Young’s modulus, which can be found by consulting
Table 10.1.
⎛ ΔL ⎞
SOLUTION Solving Equation 10.17, F = Y ⎜
⎟ A , for the change in length, we have
⎝ L0 ⎠
ΔL =
F L0
=
AY
(890 N )( 9.1 m )
(
π 0.50 × 10
−2
m
) ( 2.0 × 10
2
11
N /m
2
)
= 5.2 × 10
−4
m
______________________________________________________________________________
52. REASONING The stress in either cable is the ratio F/A of the magnitude F of the
stretching force to the cross-sectional area A of the cable. The cables have circular
cross-sections, so the area of each cable is given by A = π r 2 . Therefore, the stress in either
cable is given by
F
Stress = 2
(1)
πr
We will use Equation (1) to solve for the stretching force F2 acting on the second cable.
SOLUTION Setting the stresses in the cables equal via Equation (1), and solving for the
stretching force in the second cable gives
F2
π r22
Stress in
cable 2
F1
=
π r12
or
Stress in
cable 1
F2 =
F1r22
r12
=
( 270 N ) ( 5.1×10−3 m )
(3.5 ×10
−3
m
)
2
2
= 570 N
53. SSM REASONING AND SOLUTION According to Equation 10.20, it follows that
ΔP = − B
ΔV
− 1.0 × 10−10 m3
= − ( 2.6 × 1010 N/m 2 )
= 2.6 × 106 N/m 2
−
6
3
V0
1.0 × 10 m
(10.20)
where we have expressed the volume V0 of the cube at the ocean’s surface as
(
V0 = 1.0 ×10−2 m
)
3
= 1.0 ×10−6 m3 .
534 SIMPLE HARMONIC MOTION AND ELASTICITY
Since the pressure increases by 1.0 × 104 N/m2 per meter of depth, the depth is
2.6 × 106 N/m 2
= 260 m
2
N/m
1.0 × 104
m
______________________________________________________________________________
54. REASONING
a. According to the discussion in Section 10.8, the stress is the magnitude of the force per
unit area required to cause an elastic deformation. We can determine the maximum stress
that will fracture the femur by dividing the magnitude of the compressional force by the
cross-sectional area of the femur.
b. The strain is defined in Section 10.8 as the change in length of the femur divided by its
original length. Equation 10.17 shows how the strain ΔL/L0 is related to the stress F/A and
Young’s modulus Y (Y = 9.4 × 109 N/m2 for bone compression, according to Table 10.1).
SOLUTION
a. The maximum stress is equal to the maximum compressional force divided by the crosssectional area of the femur:
Maximum stress =
6.8 × 104 N
F
8
2
=
= 1.7 × 10 N/m
−
4
2
A 4.0 × 10 m
b. The strain ΔL/L can be found by rearranging Equation 10.17:
(
)
ΔL
1 ⎛F⎞
1
⎛
⎞
8
2
−2
= ⎜ ⎟ = ⎜
⎟ 1.7 × 10 N/m = 1.8 × 10
9
2
L0
Y ⎝ A⎠
⎝ 9.4 × 10 N/m ⎠
Stress
Strain
______________________________________________________________________________
55. REASONING The change ΔL in the length of the rope is given by ΔL = F L0 / (Y A ) (see
Equation 10.17), where F is the magnitude of the stretching force, L0 is the unstretched
length of the rope, A is its cross-sectional area, and Y is Young’s modulus for nylon (see
Table 10.1). All the variables except F are known. According to Newton’s third law, the
action-reaction law, the force exerted on the rope by the skier is equal in magnitude to the
force exerted on the skier by the rope. The force exerted on the skier by the rope can be
obtained from Newton’s second law, since the mass and acceleration of the skier are known.
Chapter 10 Problems
535
SOLUTION The change ΔL in the length of the rope is
ΔL =
F L0
YA
(10.17)
Two horizontal forces act on the skier: (1) the towing force (magnitude = F) and (2) the
resistive force (magnitude = f ) due to the water. The net force acting on the skier has a
magnitude of F − f. According to Newton’s second law (see Section 4.3), this net force is
equal to the skier’s mass m times the magnitude a of her acceleration, or
F − f =ma
Solving this equation for F and substituting the result into Equation 10.17, we find that
ΔL =
F L0
YA
=
(f
+ m a ) L0
YA
⎡130 N + ( 59 kg ) ( 0.85 m/s 2 ) ⎤(12 m )
⎦
=⎣
= 2.9 ×10−2 m
( 3.7 ×109 N/m2 )( 2.0 ×10−5 m2 )
We have taken the value of Y = 3.7 × 109 N/m2 for nylon from Table 10.1.
______________________________________________________________________________
56. REASONING The shear stress is equal to the magnitude of the shearing force exerted on
the bar divided by the cross sectional area of the bar. The vertical deflection ΔY of the right
end of the bar is given by Equation 10.18 [ F = S(ΔY / L0 ) A ].
SOLUTION
a. The stress is
F mg (160 kg)(9.80 m/s 2 )
=
=
= 4.9 × 106 N/m 2
–4
2
A
A
3.2 × 10 m
b. Taking the value for the shear modulus S of steel from Table 10.2, we find that the
vertical deflection ΔY of the right end of the bar is
0.10 m
⎛F L
Δ Y = ⎝ ⎞⎠ 0 = (4.9 × 106 N/m 2 )
= 6.0 × 10 –6 m
10
2
A S
8.1× 10 N/m
______________________________________________________________________________
57. REASONING AND SOLUTION
Equation 10.17,
The amount of compression can be obtained from
⎛ ΔL ⎞
⎟A
F = Y⎜
⎝ L0 ⎠
where F is the magnitude of the force on the stand due to the weight of the statue. Solving
for ΔL gives
536 SIMPLE HARMONIC MOTION AND ELASTICITY
(3500 kg)(9.80 m/s 2 )(1.8 m)
= 3.7 ×10 −5 m
YA
YA
(2.3 × 1010 N/m 2 )(7.3 × 10 −2 m 2 )
______________________________________________________________________________
ΔL =
FL0
=
mgL 0
=
58. REASONING AND SOLUTION F = S(ΔX/L0)A for the shearing force. The shear modulus
S for copper is given in Table 10.2. From the figure we also see that tan θ = (ΔX/L0) so that
⎡
F ⎞
6.0 × 106 N
−1 ⎢
θ = tan ⎜ ⎟ = tan
⎢ 4.2 × 1010 N/m 2 0.090 m 2
⎝ SA ⎠
⎣
−1 ⎛
(
)(
)
⎤
⎥ = 0.091°
⎥
⎦
______________________________________________________________________________
59. SSM REASONING AND SOLUTION The shearing stress is equal to the force per unit
8
area applied to the rivet. Thus, when a shearing stress of 5.0 × 10 Pa is applied to each
rivet, the force experienced by each rivet is
[
]
F = (Stress ) A = (Stress)(π r 2 ) = (5.0 × 10 8 Pa) π (5.0 × 10 –3 m) 2 = 3.9 × 10 4 N
Therefore, the maximum tension T that can be applied to each beam, assuming that each
rivet carries one-fourth of the total load, is 4 F = 1.6 × 105 N .
______________________________________________________________________________
60. REASONING Both cylinders experience the same force F. The magnitude of this force is
related to the change in length of each cylinder according to Equation 10.17:
F = Y(ΔL / L0 )A . See Table 10.1 for values of Young’s modulus Y. Each cylinder decreases
in length; the total decrease being the sum of the decreases for each cylinder.
SOLUTION The length of the copper cylinder decreases by
ΔLcopper =
FL0
FL0
(6500 N)(3.0 × 10 –2 m)
–5
=
=
= 9.0 × 10 m
2
11
2
–2
2
YA Y(π r ) (1.1× 10 N/m )π (0.25 × 10 m)
Similarly, the length of the brass decreases by
ΔLbrass =
(6500 N)(5.0 × 10 –2 m)
= 1.8 × 10 –4 m
(9.0 × 1010 N/m 2 )π (0.25 ×10 –2 m) 2
–4
Therefore, the amount by which the length of the stack decreases is 2.7 × 10 m .
______________________________________________________________________________
Chapter 10 Problems
537
61. REASONING The distance ΔX that the top surface of the disc moves relative to the bottom
surface is given by ΔX = F L0 / ( S A ) (see Equation 10.18), where F is the magnitude of the
shearing force, L0 is the thickness of the cartilage, A is the cross-sectional area, and S is the
shear modulus. Since the cross-section is circular, A = π r 2 , where r is the radius.
SOLUTION The distance (shear deformation) ΔX is
ΔX =
F L0
=
(11 N ) ( 7.0 ×10−3 m )
= 2.3 × 10−6 m
(10.18)
N/m 2 )π ( 3.0 ×10−2 m )
______________________________________________________________________________
SA
(1.2 ×107
2
62. REASONING It takes force to stretch the wire. This force arises because the tuning peg at
one end of the wire pulls on the fixed support at the other end. In accord with Newton’s
action-reaction law, the fixed support pulls back. As a result of the oppositely-directed
pulling forces at either end of the wire, the wire experiences an increased tension. For each
turn, the change in length of the wire is equal to the circumference of the tuning peg,
ΔL = 2π rp, where rp is the radius of the tuning peg. This change in length is related to the
tension by virtue of Young’s modulus Y for steel, which has a value of Y = 2.0 × 1011 N/m2.
SOLUTION The tension is the force F that is required to stretch the wire (unstrained
length = L0, cross-sectional area = A) by an amount ΔL and is determined by Young’s
modulus according to Equation 10.17:
⎛ ΔL ⎞
F = Y ⎜⎜
⎟⎟ A
⎝ L0 ⎠
Assuming that the wire has a circular cross-section, A = π rw2 , where rw is the radius of the
wire. When the tuning peg is turned through two revolutions, the length of the wire will
increase by an amount equal to twice the circumference of the peg.
Thus,
ΔL = 2(2π rp ) , where rp is the radius of the tuning peg. With these substitutions,
Equation 10.17 becomes:
⎛ 4π rp
F =Y⎜
⎜ L
⎝ 0
⎞ 2 4 Y rp
2
π rw )
⎟⎟ π rw =
(
L0
⎠
2
4(2.0 × 1011 N/m 2 )(1.8 × 10−3 m) ⎡
π (0.80 × 10−3 m) ⎤ = 1.2 × 104 N
⎣
⎦
0.76 m
______________________________________________________________________________
F=
538 SIMPLE HARMONIC MOTION AND ELASTICITY
63. REASONING AND SOLUTION The applied force, Fapplied, may be resolved into two
components, one which is parallel to the area of the top surface (a shearing force) and one
which is perpendicular to the top surface (a tensile force).
a. The change in the height of the block is caused by the component of the applied force
that is perpendicular to the top surface. Since this component is a tensile force, the change
in the height of the block can be found from Equation 10.17 with F = Fapplied(sin θ),
L0 = H0, and ΔL = ΔH:
⎛ ΔH
Fapplied (sin θ ) = Y ⎜
⎜H
⎝ 0
⎞
⎟⎟ A
⎠
From Table 10.1, the Young's modulus of copper is 1.1 × 1011 N/m2. The area, A of the top
surface is A = (5.0 × 10–2 m)(3.0 × 10–2 m) = 1.5 × 10–3 m2. Solving for ΔH gives:
ΔH =
Fapplied (sin θ ) H 0
YA
=
(1800 N)( sin 25°)(0.040 m)
= 1.8 × 10 –7 m
(1.1× 1011 N/m 2 )(1.5 × 10−3 m 2 )
b. The shear deformation of the block is caused by the component of the applied force that
is tangent to the top surface, and can be determined from Equation 10.18 with
F = Fapplied(cos θ), L0 = 0.040 m, and A = 1.5 × 10–3 m2.
⎛ ΔX ⎞
Fapplied (cos θ ) = S ⎜
A
⎜ L ⎟⎟
⎝ 0 ⎠
From Table 10.2, the shear modulus of copper is 4.2 × 1010 N/m2. Solving for ΔX gives
Fapplied (cos θ ) L0
(1800 N)( cos 25°)(0.040 m)
= 1.0 × 10−6 m
10
2
3
2
−
SA
(4.2 ×10 N/m )(1.5 ×10 m )
______________________________________________________________________________
ΔX =
=
64. REASONING The person is to be suspended by N pieces of mohair, which collectively
support the person against the downward force W = mg of gravity, where m = 75 kg is the
mass of the person and g is the magnitude of the acceleration due to gravity. Therefore, the
tension T in each piece must be T = mg N , and the total number of pieces is
N=
mg
T
The tension T in one piece of mohair is given by
(1)
Chapter 10 Problems
⎛ ΔL ⎞
⎛ ΔL ⎞ 2
πr
T =Y⎜
A=Y ⎜
⎟
⎜L ⎟
⎜ L ⎟⎟
⎝ 0⎠
⎝ 0⎠
539
(10.17)
where Y is Young’s modulus, ΔL/L0 is the strain, and A = π r2 is the cross-sectional area of a
single piece of mohair with a radius r.
SOLUTION We take the value of Young’s modulus Y = 2.9 ×109 N / m2 for mohair from
Table 10.1 in the text. Substituting Equation 10.17 into Equation (1), we obtain
(
)
( 75 kg ) 9.80 m/s 2
mg
N=
=
⎛ ΔL ⎞ 2
2.9 × 109 N/m 2 ( 0.010 ) π 31× 10−6 m
Y ⎜⎜
⎟⎟ π r
⎝ L0 ⎠
(
65. SSM
REASONING
)
(
Our approach is straightforward.
)
2
= 8400
We will begin by writing
⎡
⎛ΔL⎞ ⎤
Equation 10.17 ⎢ F = Y ⎜
A as it applies to the composite rod. In so doing, we will
⎜ L ⎟⎟ ⎥
⎝ 0 ⎠ ⎦⎥
⎣⎢
use subscripts for only those variables that have different values for the composite rod and
the aluminum and tungsten sections. Thus, we note that the force applied to the end of the
composite rod (see Figure 10.28) is also applied to each section of the rod, with the result
that the magnitude F of the force has no subscript. Similarly, the cross-sectional area A is
the same for the composite rod and for the aluminum and tungsten sections. Next, we will
express the change ΔLComposite in the length of the composite rod as the sum of the changes in
lengths of the aluminum and tungsten sections. Lastly, we will take into account that the
initial length of the composite rod is twice the initial length of either of its two sections and
thereby simply our equation algebraically to the point that we can solve it for the effective
value of Young’s modulus that applies to the composite rod.
SOLUTION Applying Equation 10.17 to the composite rod, we obtain
⎛ Δ LComposite ⎞
F = YComposite ⎜
⎟A
⎜ L0, Composite ⎟
⎝
⎠
(1)
Since the change ΔLComposite in the length of the composite rod is the sum of the changes in
lengths of the aluminum and tungsten sections, we have ΔLComposite = ΔLAluminum + ΔLTungsten.
Furthermore, the changes in length of each section can be expressed using Equation 10.17
FL0 ⎞
⎛
⎜Δ L =
⎟ , so that
YA ⎠
⎝
540 SIMPLE HARMONIC MOTION AND ELASTICITY
Δ LComposite = Δ LAluminum + Δ LTungsten =
FL0, Aluminum
YAluminum A
+
FL0, Tungsten
YTungsten A
Substituting this result into Equation (1) gives
⎛ YComposite A ⎞
⎛ YComposite A ⎞ ⎛ FL0, Aluminum FL0, Tungsten
+
F =⎜
⎟ Δ LComposite = ⎜
⎟⎜
⎜L
⎟
⎜L
⎟⎜ Y
A
YTungsten A
0,
Composite
0,
Composite
Aluminum
⎝
⎠
⎝
⎠⎝
⎛
L0, Tungsten
L0, Aluminum
+
1 = YComposite ⎜
⎜ L0, CompositeYAluminum L0, CompositeYTungsten
⎝
⎞
⎟
⎟
⎠
⎞
⎟
⎟
⎠
In this result we now use the fact that L0, Aluminum/L0, Composite = L0, Tungsten/L0, Composite = 1/2 and
obtain
⎛
⎞
1
1
1 = YComposite ⎜
+
⎟
⎜ 2YAluminum 2YTungsten ⎟
⎝
⎠
Solving for YComposite shows that
YComposite =
2YTungstenYAluminum
YTungsten + YAluminum
=
2 ( 3.6 ×1011 N/m 2 )( 6.9 ×1010 N/m 2 )
( 3.6 ×10
11
N/m ) + ( 6.9 ×10 N/m
2
10
2
)
= 1.2 × 1011 N/m 2
The values for YTungsten and YAluminum have been taken from Table 10.1.
66. REASONING AND SOLUTION From the drawing we have Δx = 3.0 × 10−3 m and
A = 2 π r Δx = 2π ( 1.00 × 10–2 m)(3.0 × 10–3 m)
We now have Stress = F/A. Therefore,
4
F = (Stress)A = (3.5 × 108 Pa)[2π(1.00 × 10–2 m)(3.0 × 10–3 m)] = 6.6 × 10 N
______________________________________________________________________________
67. REASONING AND SOLUTION
a. Strain = ΔL/L0 = F/(YA). In this case the area subjected to the compression is given by
(
)
(
⎡
2
− rin2 = π ⎢ 1.00 × 10−2
A = π rout
⎣
) − ( 4.0 ×10−3 ) ⎤⎥⎦ = 2.64 ×10−4 m2
2
2
Chapter 10 Problems
541
and the force is F = mg. Taking the value for Young’s modulus Y for bone compression
from Table 10.1, we find that
Strain =
( 63 kg ) ( 9.80 m/s 2 )
(9.4 × 10
9
)(
Pa 2.64 × 10
−4
m
2
)
= 2.5 × 10−4
b. ΔL = Strain × L0 = (2.5 × 10–4)(0.30 m) = 7.5 × 10−5 m
______________________________________________________________________________
68. REASONING We will use energy conservation. The person falls from rest and does not
rotate, so initially he has only gravitational potential energy. Ignoring air resistance and
friction, we may apply the conservation of mechanical energy. Since the person strikes the
ground stiff-legged and comes to a halt without rotating, all of the energy he had to begin
with must be absorbed by his leg as elastic potential energy. The height through which he
falls determines the amount of his gravitational potential energy and, hence, the amount of
energy his leg must absorb.
According to Equation 10.13, the elastic potential energy of an ideal spring is
PE Elastic = 12 kEffective x 2 , so we will need kEffective for a thigh bone. To find it, we consider
Equation 10.1 for the applied force needed to change the length of an ideal spring:
FxApplied = keffectivex, where keffective is the spring constant and x is the displacement. To
change the length of a bone, in comparison, the necessary applied force is given by
Equation 10.17 as follows:
⎛ΔL⎞
FxApplied = Y ⎜
A=
⎜ L ⎟⎟
⎝ 0 ⎠
(YA / L )
0
ΔL
The effective The change in
spring constant length or the
displacement x
kEffective
In this expression Y is Young’s modulus, A is the effective cross-sectional area of the bone,
L0 is the initial length of the bone, and ΔL is the change in length. Associating ΔL with x,
we see that the effective spring constant of the bone is given by
kEffective =
YA
L0
(1)
Equation 10.13, which specifies the elastic potential energy of an ideal spring as
PE Elastic = 12 kEffective x 2 , contains the displacement x of the spring. To eliminate it, we turn to
Equation 10.1, which gives the applied force as FxApplied = kEffectivex. Solving Equation 10.1
for x and substituting the result into Equation 10.13 gives
542 SIMPLE HARMONIC MOTION AND ELASTICITY
PE Elastic
⎛ F Applied
= 12 kEffective x 2 = 12 kEffective ⎜ x
⎜k
⎝ Effective
⎞ ( FxApplied )
⎟⎟ =
2kEffective
⎠
2
2
(2)
SOLUTION The conservation of mechanical energy states that the final total mechanical
energy Ef is equal to the initial total mechanical energy E0. The expression for the total
mechanical energy for a spring/object system is given by Equation 10.14, so we have
1
2
mvf2 + 12 Iω f2 + mghf + 12 kxf2 = 12 mv02 + 12 Iω 02 + mgh0 + 12 kx02
Ef
E0
Since the person does not rotate, the angular speeds ωf and ω0 are zero. The person is at rest
both initially and finally, so the initial and final translational speeds v0 and vf are also zero.
Moreover, the thighbone is initially unstrained, with the result that x0 is zero. Thus, the
above expression can be simplified to give
mghf + 12 kxf2 = mgh0
Using Equation (2) to express the final elastic potential energy of the thighbone, we can
write the simplified energy-conservation equation as follows:
mghf
(F
+
)
Applied 2
x
2kEffective
= mgh0
or
h0 − hf =
(F
)
Applied 2
x
2mgkEffective
As the man falls, his center of gravity moves from its initial height of h0 to its final height of
hf, which is a distance of h0 − hf. Using Equation (1) for the effective spring constant of the
bone, we find
h0 − hf =
(F
)
Applied 2
x
2mgkEffective
(F
=
)
Applied 2
x
L0
2mgYA
( 7.0 ×10 N ) ( 0.55 m )
=
2 ( 65 kg ) ( 9.80 m/s )( 9.4 × 10 N/m )( 4.0 × 10
4
2
2
9
2
−4
m2 )
= 0.56 m
69. SSM WWW REASONING The strain in the wire is given by ΔL / L0 . From Equation
10.17, the strain is therefore given by
ΔL F
=
L0 YA
(1)
Chapter 10 Problems
543
where F must be equal to the magnitude of the centripetal force that keeps the stone moving
in the circular path of radius R. Table 10.1 gives the value of Y for steel.
SOLUTION Combining Equation (1) with Equation 5.3 for the magnitude of the
centripetal force, we obtain
F
(mv 2 / R)
(8.0 kg)(12 m/s) 2 /(4.0 m)
ΔL
=
=
=
= 4.6 × 10 –4
L0 Y ( π r 2 ) Y (π r 2 ) (2.0 ×10 11 Pa) π (1.0 × 10 –3 m) 2
______________________________________________________________________________
70. REASONING We will find displacement ΔX of the upper surface relative to the lower
⎛ ΔX ⎞
surface from F = S ⎜
A (Equation 10.18), where F is the applied horizontal force (see
⎜ L ⎟⎟
⎝ 0 ⎠
Figure 10.31), S is the shear modulus of brass, L0 is the distance between the upper and
lower surfaces, and A is the cross-sectional area of either surface. Solving Equation 10.18
for ΔX yields
FL
⎛ L0 ⎞
⎛F⎞
(1)
ΔX = 0 = ⎜ ⎟ ×
⎜ ⎟
SA
⎝S⎠
⎝ A⎠
Same for all
orientations
Depends upon
orientation
Note that the term L0/A in Equation (1) depends on the orientation of the block. The block
can rest on any one of its three unique surfaces. To make the displacement ΔX of the top
surface as large as possible, Equation (1) shows that we must orient the block so as to make
the ratio L0/A as large as possible. The largest possible value for the numerator of this ratio
is the longest side of the block, so we choose L0 = 0.040 m. Gluing the block to the table so
that the four longest edges are vertical means that the
F
upper surface of the block is the surface with the
smallest area: A = 0.010 m × 0.020 m (see the drawing).
This combination of the largest possible value of L0 and
L0 = 0.040 m
the smallest possible value of A makes the ratio L0/A as
large as possible. Therefore, this is the orientation with
the greatest possible displacement ΔX of the upper
0.010 m
surface relative to the lower surface when the horizontal
0.020 m
force is applied to the upper surface.
SOLUTION We obtain the shear modulus S = 3.5 ×1010 N/m 2 of brass from Table 10.2 in
the text. From Equation (1), the maximum possible displacement of the upper surface
relative to the lower surface is
ΔX =
FL0
SA
=
( 770 N )( 0.040 m )
=
10
2
×
3.5
10
N/m
0.010
m
0.020
m
(
)(
)
(
)
4.4 × 10−6 m
544 SIMPLE HARMONIC MOTION AND ELASTICITY
71. REASONING The unstretched length L0 of the cable can be found from the relation
L0 = Y A ( ΔL ) / F (Equation 10.17), where Y is Young’s modulus for steel (see Table 10.1),
A is the cross-sectional area of the cable, ΔL is the amount by which it stretches, and F is the
magnitude of the stretching force. All the variables except F are known. According to
Newton’s third law, the action-reaction law, the force exerted on the cable by the skier has
the same magnitude as the force exerted on the skier by the cable. The force exerted on the
skier by the cable can be obtained from Newton’s second law, since the mass and
acceleration of the skier are known.
SOLUTION The unstrained length of the cable is
L0 =
Y A ( ΔL )
F
(10.17)
To determine F, we examine the following free-body diagram of the skier. For convenience,
the +x direction is taken to be parallel to the slope and to point upward (see the drawing).
F
+x
mg sin 12°
f
W = mg
W = mg
12°
12°
Free-body diagram for the skier
Three forces act on the skier in the x direction: (1) the towing force (magnitude = F), (2) the
frictional force (magnitude = f ) exerted on the skis by the snow, and (3) the component of
the skier’s weight that is parallel to the x axis (magnitude = W sin12° = mg sin 12°). This
component is shown to the right of the free-body diagram. The net force acting on the skier
has a magnitude of F − f − mg sin12° . According to Newton’s second law (see Section 4.3),
this net force is equal to the skier’s mass times the magnitude of her acceleration, or
F − f − mg sin12° = m a
Solving this equation for F and substituting the result into Equation 10.17, we find that
L0 =
( 2.0 ×1011 N/m 2 )( 7.8 ×10−5 m 2 )( 2.0 × 10−4 m )
Y A ( ΔL )
=
= 12 m
m a + f + mg sin12° ( 61 kg ) (1.1 m/s 2 ) + ( 68 N ) + ( 61 kg ) ( 9.80 m/s 2 ) sin12°
We have taken the value of Y = 2.0 × 1011 N/m2 for steel from Table 10.1.
Chapter 10 Problems
545
______________________________________________________________________________
72. REASONING The unstrained length L0 of the cord is related to the stretching force
(tension) F on the cord, the amount ΔL by which the cord is stretched, the cord’s
cross-sectional area A, and Young’s modulus Y = 3.7 × 109 N/m2 (see Table 10.1 in the text)
by
⎛ ΔL ⎞
(10.17)
F =Y⎜
A
⎜ L ⎟⎟
⎝ 0⎠
While swinging at a speed v on the end of the cord, the bowling ball (mass = m) follows a
circular path, and the tension F in the cord must be sufficient to supply the necessary
mv 2
centripetal force Fc =
(Equation 5.3). Ignoring the stretch in the cord, the radius r of
r
the circle that the bowling ball traverses is equal to the length L0 of the cord. Thus,
Equation 5.3 becomes
mv 2
(5.3)
Fc =
L0
At the lowest point, the upward force F on the ball is opposed by the downward weight
force mg, so the difference between these two forces equals the centripetal force Fc:
Fc =
mv 2
= F − mg
L0
or
F=
mv 2
+ mg
L0
(1)
To find the speed of the bowling ball, we take advantage of energy conservation. As the ball
swings downward from release to its lowest point, it is subject to only one nonconservative
force: the tension in the cord. This force is perpendicular to the bowling ball’s velocity at
every instant, and, therefore, does no work. Consequently, the total mechanical energy
E = 12 mv 2 + 12 I ω 2 + mgh + 12 kx 2 (Equation 10.14) of the bowling ball is conserved. The
bowling ball starts from rest and undergoes no rotation (ω0 = ωf = 0 rad/s), so its initial
translational kinetic energy is zero, as are its initial and final rotational kinetic energies. The
cord is initially unstretched, and its final stretch x = ΔL is negligible, so the initial and final
elastic potential energies are both zero. When we apply the energy conservation principle,
then, we obtain
1 mv 2 + mgh = mgh
(2)
f
f
0
2
E
f
E0
546 SIMPLE HARMONIC MOTION AND ELASTICITY
Solving Equation (2) for the square of the speed vf2 of the bowling ball at its lowest point,
we find that
1
2
m vf2 + m ghf = m gh0
vf2 = 2 g ( h0 − hf )
or
(3)
We will use Equations (10.17), (1), and (3) to solve for the unstretched length L0 of the cord.
SOLUTION Because Equation (1) refers to the instant when the bowling ball reaches its
lowest point, the speed v in Equation (1) is identical with the final speed vf in Equation (3):
v = vf. Making this substitution in Equation (1), and setting Equation (10.17) and
Equation (1) equal, we obtain
⎛ ΔL ⎞
mv 2
Y⎜
A = f + mg
⎟
⎜L ⎟
L0
⎝ 0⎠
(4)
Multiplying both sides of Equation (4) by L0 and solving, we find
⎛ ΔL ⎞
mvf2
⎜
⎟
A L0 =
L + mgL0
Y
⎜ L0 ⎟
L0 0
⎝
⎠
or
Y ( ΔL ) A − mvf2
L0 =
mg
(5)
Substituting Equation (3) for the square of the final speed into Equation (5) yields the
unstrained length of the cord:
L0 =
Y ( ΔL ) A − 2mg ( h0 − hf )
mg
=
Y ( ΔL ) A
− 2 ( h0 − hf )
mg
3.7 × 109 N/m 2 )( 2.7 × 10 −3 m )( 3.4 × 10−5 m 2 )
(
=
− 2 (1.4 m ) = 2.3 m
( 6.8 kg ) ( 9.80 m/s 2 )
73. SSM REASONING The change in length of the wire is, According to Equation 10.17,
ΔL = FL0 / YA , where the force F is equal to the tension T in the wire. The tension in the
wire can be found by applying Newton's second law to the two crates.
Chapter 10 Problems
SOLUTION The drawing shows the free-body diagrams for
the two crates. Taking up as the positive direction, Newton's
second law for each of the two crates gives
T − m1g = m1a
(1)
T − m2 g = –m 2a
(2)
T
547
T
m1g
m2g
⎛ T − m2g ⎞
⎟ . Substituting into Equation (1) gives
Solving Equation (2) for a, we find a = −⎜
⎝ m2 ⎠
m
T − 2 m1g + 1 T = 0
m2
Solving for T we find
T=
2m1m2 g 2(3.0 kg)(5.0 kg)(9.80 m/s2 )
=
= 37 N
3.0 kg + 5.0 kg
m1 + m2
Using the value given in Table 10.1 for Young’s modulus Y of steel, we find, therefore, that
the change in length of the wire is given by Equation 10.17 as
ΔL =
(37 N)(1.5 m)
(2.0 × 1011 N/m 2 )(1.3 × 10 –5 m 2 )
= 2.1 × 10–5 m
______________________________________________________________________________
74. REASONING The number of times the diaphragm moves back and forth is the frequency f
of the motion (in cycles/s or Hz) times the time interval t. The frequency f is related to the
angular frequency ω (in rad/s) by Equation 10.6 (ω = 2πf ).
SOLUTION Solving Equation 10.6 (ω = 2πf ) for the frequency f gives
f =
ω
2π
The number of times the diaphragm moves back and forth in 2.5 s is
⎛ω
Number of times = ft = ⎜
⎝ 2π
⎞ ⎛ 7.54 ×104 rad/s ⎞
4
⎟ ( 2.5 s ) = 3.00 × 10
⎟t = ⎜
2π
⎠ ⎝
⎠
548 SIMPLE HARMONIC MOTION AND ELASTICITY
75. REASONING AND SOLUTION From Conceptual Example 2, we know that when the
spring is cut in half, the spring constant for each half is twice as large as that of the original
spring. In this case, the spring is cut into four shorter springs. Thus, each of the four shorter
springs with 25 coils has a spring constant of 4 × 420 N/m = 1680 N/m .
The angular frequency of simple harmonic motion is given by Equation 10.11:
k
1680 N/m
=
= 6.0 rad/s
m
46 kg
______________________________________________________________________________
ω=
76. REASONING AND SOLUTION Applying Equation 10.16 and recalling that frequency
and period are related by f = 1/T,
2π
g
2π f =
=
T
L
where L is the length of the pendulum. Thus,
T = 2π
L
g
Solving for L gives
2
2
⎛ T ⎞
2 ⎛ 9.2 s ⎞
L = g⎜
⎟ = ( 9.80 m/s ) ⎜
⎟ = 21 m
⎝ 2π ⎠
⎝ 2π ⎠
______________________________________________________________________________
77.
SSM WWW REASONING Each spring supports one-quarter of the total mass mtotal of
the system (the empty car plus the four passengers), or
one passenger is equal to
1m
4 total
1m
.
4 total
The mass mone passenger of
minus one-quarter of the mass mempty car of the empty car:
mone passenger = 14 mtotal − 14 mempty car
(1)
The mass of the empty car is known. Since the car and its passengers oscillate up and down
in simple harmonic motion, the angular frequency ω of oscillation is related to the spring
constant k and the mass 14 mtotal supported by each spring by:
ω=
Solving this expression for
Equation (1) gives
1m
4 total
k
1m
4 total
( 14 mtotal = k / ω 2 ) and substituting the result into
(10.11)
Chapter 10 Problems
mone passenger =
k
ω
2
− 14 mempty car
549
(2)
The angular frequency ω is inversely related to the period T of oscillation by ω = 2π / T (see
Equation 10.4). Substituting this expression for ω into Equation (2) yields
mone passenger =
k
⎛ 2π ⎞
⎜
⎟
⎝ T ⎠
2
− 14 mempty car
SOLUTION The mass of one of the passengers is
mone passenger =
k
2
− 14 mempty car =
1.30 ×105 N/m
2
− 14 (1560 kg ) = 61 kg
⎛ 2π ⎞
⎛ 2π ⎞
⎜
⎟
⎜
⎟
⎝ T ⎠
⎝ 0.370 s ⎠
______________________________________________________________________________
78. REASONING The amount ΔL by which the bone changes length when a compression
FL
force or a tension force acts on it is specified by Δ L = 0 (Equation 10.17), where F
YA
denotes the magnitude of either type of force, L0 is the initial length of the bone, Y is the
appropriate Young’s modulus, and A is the cross-sectional area of the bone. The values of
Young’s modulus are given in Table 10.1 (YCompression = 9.4 × 109 N/m2 and
YTension = 1.6 × 1010 N/m2). The values for F, L0, and A are not given, but it is important to
recognize that these variables have the same values for both types of forces. We will apply
Equation 10.17 twice, once for the compression force and once for the tension force. Since
F, L0, and A have the same values in both of the resulting equations, we will be able to
eliminate them algebraically and determine the amount ΔLTension by which the bone stretches.
SOLUTION Applying Equation 10.17 for both types of forces gives
Δ LTension =
FL0
and
YTension A
Δ LCompression =
FL0
YCompression A
Dividing the left-hand equation by the right-hand equation and eliminating the common
variables algebraically shows that
FL0
Δ LTension
Δ LCompression
=
YCompression
YTension A
=
FL0
YTension
YCompression A
550 SIMPLE HARMONIC MOTION AND ELASTICITY
Solving for ΔLTension, we find that
⎛ YCompression ⎞
⎛ 9.4 × 109 N/m 2 ⎞
Δ LTension = ⎜
2.7 × 10−5 m ) = 1.6 × 10−5 m
⎟⎟ Δ LCompression = ⎜
10
2 ⎟(
⎜ Y
⎝ 1.6 × 10 N/m ⎠
⎝ Tension ⎠
______________________________________________________________________________
79. REASONING The applied force required to stretch the bow string is given by Equation
10.1 as FxApplied = k x , where k is the spring constant and x is the displacement of the string.
Solving for the displacement gives x = FxApplied / k .
SOLUTION The displacement of the bow string is
x=
FxApplied
k
=
+240 N
= +0.50 m
480 N/m
____________________________________________________________________________________________
80. REASONING AND SOLUTION Equation 10.20 gives the desired result. Solving for
Δ V / V0 and taking the value for the bulk modulus B of aluminum from Table 10.3, we
obtain
ΔP
− 1.01 × 105 Pa
ΔV
–6
=−
=−
10
2 = 1.4 × 10
B
V0
7.1× 10 N/m
______________________________________________________________________________
81. REASONING AND SOLUTION
a. Now look at conservation of energy before and after the split
Before
1 mv 2
max
2
= 12 kA2
Solving for the amplitude A gives
A = vmax
After
1⎛
2⎜
⎝
m
k
m⎞ 2 1⎛m⎞ 2
2
1
⎟ v′ = 2 ⎜ ⎟ vmax = 2 kA′
2⎠
⎝2⎠
Solving for the amplitude A' gives
A′ = vmax
Therefore, we find that
m
2k
Chapter 10 Problems
A' =
551
A 5.08 ×10−2 m
=
= 3.59 ×10−2 m
2
2
Similarly, for the frequency, we can show that
f' = f
2 = (3.00 Hz) 2 = 4.24 Hz
b. If the block splits at one of the extreme positions, the amplitude of the SHM would not
change, so it would remain as 5.08 × 10 –2 m
The frequency would be
f' = f
2 = (3.00 Hz)
2 = 4.24 Hz
______________________________________________________________________________
82. REASONING
a. The acceleration of the box is zero when the net force acting on it is zero, in accord with
Newton’s second law of motion. The net force includes the box’s weight (directed
downward) and the restoring force of the spring (directed upward). The condition that the
net force is zero will allow us to determine the magnitude of the spring’s displacement.
b. The speed of the box is zero when the spring is fully compressed, but the acceleration of
the box is not zero at this instant. If the acceleration were zero, the box would be in
equilibrium, and the net force on it would be zero. However, the box accelerates upward
because the spring is exerting an upward force that is greater than the downward force due
to the weight of the box. Thus, we cannot proceed as in part (a), so instead we will use
energy conservation to determine the magnitude of the spring’s displacement.
SOLUTION
a. The drawing at the right shows the two forces acting on the box: its weight mg
and the restoring force Fy exerted by the spring. At the instant the acceleration of
the box is zero, it is in equilibrium. According to Equation 4.9b, the net force ΣFy in
the y direction must be zero, ΣFy = 0.
+y
The restoring force is given by Equation 10.2 as Fy = –ky, where k is the spring
constant and y is the displacement of the spring (assumed to be in the downward, or
negative, direction). Thus, the condition for equilibrium can be written as
mg
− ky + mg = 0
or
y=−
mg
k
ΣFy
Solving for the magnitude of the spring’s displacement gives
Fy
552 SIMPLE HARMONIC MOTION AND ELASTICITY
(
1.5 kg ) 9.80 m /s
Magnitude of spring's = mg = (
displacement
k
450 N /m
2
)=
3.3 × 10−2 m
b. The conservation of mechanical energy states that the final total mechanical energy Ef is
equal to the initial total mechanical energy E0, or Ef = E0 (Equation 6.9a). The expression
for the total mechanical energy of an object is given by Equation 10.14. Thus, the
conservation of total mechanical energy can be written as
1
2
m vf2 + 12 I ω f2 + m g hf + 12 k yf2 =
1
2
m v02 + 12 I ω 02 + m g h0 + 12 k y02
Ef
E0
We can simplify this equation by noting which variables are zero. Since the box comes to a
momentary halt, vf = 0 m/s. The box does not rotate, so its angular speed is zero, ωf = ω0 =
0 rad/s. Initially, the spring is unstretched, so that y0 = 0 m. Setting these terms equal to zero
in the equation above gives
2
2
mghf + 12 k yf = 12 mv0 + mgh0
The vertical displacement hf – h0 through which the box falls is equal to the displacement yf
of the spring, so yf = hf – h0. Note that yf is negative, because hf is less than h0. The
downward-moving box compresses the spring in the downward direction, which, as usual,
we take to be the negative direction. Substituting this expression for yf into the equation
above and rearranging terms, we find that
1 mv 2
0
2
− mg ( hf − h0 ) −
1 k y2
f
2
=0
y
f
or
1 k y2
f
2
2
+ mg yf − 12 mv0 = 0
This is a quadratic equation in the variable yf. The solution is
yf =
−mg ±
( mg )2 − 4 ( 12 k ) ( − 12 mv02 )
2
( 12 k )
Substituting in the numbers, we find that
2
=
− mg
⎛ mg ⎞ ⎛ m ⎞ 2
± ⎜
⎟ + ⎜ ⎟ v0
k
⎝ k ⎠ ⎝k⎠
Chapter 10 Problems
(
)
− (1.5 kg ) 9.80 m/s 2
yf =
±
450 N/m
= 1.1 × 10−2 m
or
(
)
553
2
⎡ (1.5 kg ) 9.80 m/s 2 ⎤
⎛ 1.5 kg ⎞
2
⎢
⎥ +⎜
⎟ ( 0.49 m/s )
450 N/m
⎣⎢
⎦⎥
⎝ 450 N/m ⎠
− 7.6 × 10−2 m
The positive answer is discarded because the spring is compressed downward by the falling
box, so the displacement of the spring is negative.
Therefore, the magnitude of the spring’s displacement is 7.6 × 10−2 m .
83. REASONING
⎛
k ⎞
a. The angular frequency ω (in rad/s) is given by Equation 10.11 ⎜⎜ ω =
⎟ , where k is the
m ⎟⎠
⎝
spring constant and m is the mass of the object. The frequency f (in Hz) can be obtained
from the angular frequency by using Equation 10.6 (ω = 2πf).
b. The block loses contact with the spring when the amplitude of the oscillation is
sufficiently large. To understand why, consider the block at the very top of its oscillation
cycle. There it is accelerating downward, with the maximum acceleration amax of simple
harmonic motion. Contact is maintained with the spring, as long as the magnitude of this
acceleration is less than the magnitude g of the acceleration due to gravity. If amax is greater
than g, the end of the spring falls away from under the block. amax is given by Equation
10.10 ( amax = Aω 2 ) , from which we can obtain the amplitude A when amax = g.
SOLUTION
a. From Equations 10.6 and 10.11 we have
ω = 2π f =
k
m
or
f =
1
2π
k
1
=
m 2π
112 N/m
= 2.66 Hz
0.400 kg
b. Using Equation 10.10 with amax = g gives
amax = g = Aω 2
⎛
Substituting Equation 10.11 ⎜⎜ ω =
⎝
or
A=
g
ω2
k ⎞
⎟ into this result gives
m ⎟⎠
554 SIMPLE HARMONIC MOTION AND ELASTICITY
A=
g
ω2
=
(
g
k /m
)
2
2
gm ( 9.80 m/s ) ( 0.400 kg )
=
=
= 0.0350 m
k
112 N/m
84. REASONING As the climber falls, only two forces act on him: his weight and the elastic
force of the nylon rope. Both of these forces are conservative forces, so the falling climber
obeys the conservation of mechanical energy. We will use this conservation law to
determine how much the rope is stretched when it breaks his fall and momentarily brings
him to rest.
SOLUTION
The conservation of mechanical energy states that the final total mechanical energy Ef is
equal to the initial total mechanical energy E0, or Ef = E0 (Equation 6.9a). The expression for
the total mechanical energy of an object oscillating on a spring is given by Equation 10.14.
Thus, the conservation of total mechanical energy can be written as
1
2
m vf2 + 12 I ωf2 + m g hf + 12 k yf2 =
1
2
m v02 + 12 I ω02 + m g h0 + 12 k y02
Ef
E0
Before going any further, let’s simplify this equation by noting which variables are zero.
Since the climber starts and ends at rest, vf = v0 = 0 m/s. The climber does not rotate, so his
angular speed is zero, ωf = ω0 = 0 rad/s. Initially, the spring is unstretched, so that y0 = 0 m.
Setting these terms to zero in the equation above gives
m g hf + 12 k yf2 = m g h0
Rearranging the terms in this equation, we have
1 ky 2
2 f
+ mg
( hf − h0 )
=0
y − 0.750 m
f
Note that the climber falls a distance of 0.750 m before the rope starts to stretch, and yf is the
displacement of the stretched rope. Since yf points downward, it is considered to be
negative. Thus, yf – 0.750 m is the total downward displacement of the falling climber,
which is also equal to hf – h0. With this substitution, the equation above becomes
1 ky 2
2 f
+ mgyf − mg ( 0.750 m ) = 0
This is a quadratic equation in the variable yf. Using the quadratic formula gives
Chapter 10 Problems
yf =
yf =
− mg ±
555
( mg )2 − 4 ( 12 k ) ⎡⎣ − mg ( 0.750 m )⎤⎦
2
(
( 12 k )
− ( 86.0 kg ) 9.80 m /s
2
)
3
1.20 × 10 N /m
±
(
)
2
⎡( 86.0 kg ) 9.80 m /s 2 ⎤ − 4
⎣⎢
⎦⎥
( 12 ) (1.20 × 103 N /m ) ⎡⎣⎢− (86.0 kg ) ( 9.80 m /s2 ) ( 0.750 m )⎤⎦⎥
3
1.20 × 10 N /m
There are two answers, yf = +0.54 m and –1.95 m. Since the rope is stretched in the
downward direction, which we have taken to be the negative direction, the displacement is
–1.95 m. Thus, the amount that the rope is stretched is 1.95 m .
______________________________________________________________________________
85. REASONING The two blocks and the spring between them constitute the system in this
problem. Since the surface is frictionless and the weights of the blocks are balanced by the
normal forces from the surface, no net external force acts on the system. Thus, the system’s
total mechanical energy and total linear momentum are each conserved. The conservation
of these two quantities will give us two equations containing the two unknown speeds with
which the blocks move away. Using these equations, we will be able to obtain the speeds.
SOLUTION The conservation of mechanical energy states that the final total mechanical
energy Ef is equal to the initial total mechanical energy E0. Only the translational kinetic
energies of the blocks and the elastic potential energy of the spring are of interest here.
There is no rotational kinetic energy since there is no rotation. Gravitational potential
energy plays no role, because the surface is horizontal and the vertical height does not
change. Thus, the expression for the conservation of the total mechanical energy is
1
2
2
2
m1vf12 + 12 m2 vf22 + 12 kxf2 = 12 m1v01
+ 12 m2 v02
+ 12 kx02
Ef
E0
Since the blocks are initially at rest, the initial translational speeds v01 and v02 are zero. In
addition, the spring is neither compressed nor stretched after it is released, so that xf = 0 m.
Thus, the above expression can be simplified as follows:
1
2
m1vf12 + 12 m2 vf22 = 12 kx02
(1)
556 SIMPLE HARMONIC MOTION AND ELASTICITY
Remembering that linear momentum is mass times velocity, we can express the
conservation of the total linear momentum of the system as follows:
m1vf1 + m2 vf2 = m1v01 + m2 v02
Final total momentum
Initial total momentum
Since the blocks are initially at rest, v01 and v02 are zero, so that the expression for the
conservation of linear momentum becomes
m1vf1 + m2vf2 = 0
or
vf2 = −
m1vf1
(2)
m2
Substituting this result into Equation (1) gives
2
1
2
⎛ mv ⎞
m v + m2 ⎜⎜ − 1 f1 ⎟⎟ = 12 kx02
⎝ m2 ⎠
2
1 f1
1
2
Solving for vf1, which we define to be the final speed of the 11.2-kg block, shows that
vf1 =
m2 kx02
m1 ( m2 + m1 )
=
( 21.7 kg )(1330 N/m )( 0.141 m )
(11.2 kg )( 21.7 kg + 11.2 kg )
2
= 1.25 m/s
Substituting this value into Equation (2) gives
vf2 = −
m1vf1
(11.2 kg )(1.25 m/s ) = −0.645 m/s
=−
m2
21.7 kg
The speed of the 21.7-kg block is the magnitude of this result or 0.645 m/s .
86. REASONING AND SOLUTION Strain = ΔL/L0 = F/(YA) where F = mg and A = π r2.
Setting the strain for the spider thread equal to the strain for the wire
F
F′
=
YA
Y ′ A′
Spider
thread
so that
Aluminum
wire
Thus,
r ′2 =
F ′Y r 2
F Y′
F
F′
=
2
Yr
Y ′ r ′2
Chapter 10 Problems
557
Taking the value for Young’s modulus Y ′ for aluminum from Table 10.1, we find that
r′ =
( 95 kg ) ( 9.80 m/s 2 )( 4.5 × 109 Pa )(13 × 10−6 m )
(1.0 × 10
−3
)(
kg 9.80 m/s
2
)( 6.9 × 10
10
Pa
2
)
= 1.0 × 10−3 m
______________________________________________________________________________
87. SSM REASONING AND SOLUTION The natural frequency of the suspension system
is given by Equation 10.11:
k
1.50 × 106 N/m
=
= 83.5 rad/s
ω=
m
215 kg
Thus, the wheel will resonate when its angular speed is 83.5 rad/s. This corresponds to a
linear speed of
v = rω = (0.400 m)(83.5 rad/s) = 33.4 m/s
______________________________________________________________________________
88. REASONING AND SOLUTION The bullet (mass m) moves with speed v, strikes the
block (mass M) in an inelastic collision and the two move together with a final speed V. We
first need to employ the conservation of linear momentum to the collision to obtain an
expression for the final speed:
mv = ( m + M ) V
or
V =
mv
m+M
The block/bullet system now compresses the spring by an amount x. During the
compression the total mechanical energy is conserved so that
1
2
( m + M )V 2
= 12 kx 2
Substituting the expression for V into this equation, we obtain
2
mv ⎞
= 12 kx 2
( m + M ) ⎛⎜
⎟
⎝m+ M ⎠
Solving this expression for v gives
1
2
k x2 ( m + M )
v=
=
m2
(845 N/m )( 0.200 m ) ( 2.51 kg )
( 0.0100 kg )2
2
= 921 m/s
______________________________________________________________________________
558 SIMPLE HARMONIC MOTION AND ELASTICITY
89. SSM WWW REASONING Equation 10.20 can be used to find the fractional change
in volume of the brass sphere when it is exposed to the Venusian atmosphere. Once the
fractional change in volume is known, it can be used to calculate the fractional change in
radius.
SOLUTION According to Equation 10.20, the fractional change in volume is
ΔP
8.9 × 106 Pa
ΔV
–4
=−
=−
= −1.33 × 10
10
B
V0
6.7 × 10 Pa
Here, we have used the fact that ΔP = 9.0 ×106 Pa − 1.0 ×105 Pa = 8.9 × 106 Pa , and we have
taken the value for the bulk modulus B of brass from Table 10.3. The initial volume of the
sphere is V0 = 43 π r03 . If we assume that the change in the radius of the sphere is very small
relative to the initial radius, we can think of the sphere's change in volume as the addition or
subtraction of a spherical shell of volume ΔV , whose radius is r0 and whose thickness is
Δr . Then, the change in volume of the sphere is equal to the volume of the shell and is
given by ΔV = 4π r02 Δr . Combining the expressions for V0 and ΔV , and solving for
(
)
Δr / r0 , we have
Δr ⎛ 1 ⎞ ΔV
=⎜ ⎟
r0 ⎝ 3 ⎠ V0
Therefore,
(
)
Δr 1
= −1.33 ×10−4 = −4.4 × 10−5
r0 3
______________________________________________________________________________
90. REASONING AND SOLUTION The frequency f of the simple harmonic motion is given
by Equations 10.6 and 10.11 as f = (1/ 2π ) k / m . If we compare Equation 10.17, which
governs the stretching and compression of a solid rod with Equation 10.1, we find that x is
analogous to ΔL and k is analogous to the term YA/L0:
F=
YA
ΔL
L0 x
k
The value of Young’s modulus for copper is given in Table 10.1. Assuming that the rod has
a circular cross-section, its area A is equal to π r2, and we have
Chapter 10 Problems
1
f =
2π
1
=
2π
1
k
=
m 2π
1
YA
=
L0 m 2π
559
Y (π r 2 )
L0 m
(1.1 × 1011 N/m2 ) π ( 3.0 × 10−3 m )
( 2.0 m )( 9.0 kg )
2
= 66 Hz
______________________________________________________________________________
91. SSM
REASONING The angular frequency for simple harmonic motion is given by
Equation 10.11 as ω = k / m . Since the frequency f is related to the angular frequency ω
by f = ω/(2π) and f is related to the period T by f = 1/T, the period of the motion is given by
T=
2π
ω
= 2π
k
m
SOLUTION
a. When m1 = m2 = 3.0 kg, we have that
3.0 kg
= 0.99 s
120 N/m
T1 = T2 = 2π
Both particles will pass through the position x = 0 m for the first time one-quarter of the way
through one cycle, or
0.99 s
Δt = 14 T1 = 14 T2 =
= 0.25 s
4
b. T1 = 0.99 s, as in part (a) above, while
27.0 kg
= 3.0 s
120 N/m
Each particle will pass through the position x = 0 m every odd-quarter of a cycle,
3
5
1
4 T , 4 T , 4 T , ... Thus, the two particles will pass through x = 0 m when
T2 = 2π
3.0-kg particle
t = 14 T1 , 34 T1 , 54 T1 , …
27.0-kg particle
t = 14 T2 , 34 T2 , 54 T2 , …
Since T2 = 3T1, we see that both particles will be at x = 0 m simultaneously when
t = 34 T1 , or t = 14 T2 = 34 T1. Thus,
t = 34 T1 =
3
4
( 0.99 s ) =
0.75 s
560 SIMPLE HARMONIC MOTION AND ELASTICITY
______________________________________________________________________________
92. REASONING If we compare Equation 10.17, which governs the stretching and
compression of a solid cylinder, with Equation 10.1, we find that x is analogous to ΔL and k
is analogous to the term YA/L0:
⎛ YA ⎞
F = ⎜ ⎟ ΔL
⎝ L0 ⎠ x
k
SOLUTION
a. Solving for k we have
YA Y (π r 2 ) ( 3.1 × 106 N/m 2 )π ( 0.091 × 10−2 m )
k=
=
=
= 3.2 × 102 N/m
−2
L0
L0
2.5 × 10 m
2
b. The work done by the variable force is equal to the area under the F-versus-x curve. The
amount x of stretch is
F
3.0 × 10−2 N
x= =
= 9.4 × 10−5 m
2
k 3.2 × 10 N/m
The work done is
W = 12 F x =
1
2
( 3.0 × 10−2 N )( 9.4 × 10−5 m ) = 1.4 × 10−6 J
______________________________________________________________________________