Physics 1010:
The Physics of Everyday Life
TODAY
• Sound
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SOUND AND MUSIC
• Sound propagates as a wave (in air, liquids,
solids)
• Sound is compressional (longitudinal) wave
• Waves are characterized by speed,
frequency, and period
• Speed, frequency, and period are related
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Question:
P You are at the ball park sitting in the bleachers in the
outfield (~325 ft from the batter). You see the bat hit
the ball. About how long will it take before you hear the
bat hit the ball?
a.
b.
c.
d.
About 30 seconds
About 3 seconds
About 0.3 seconds
There will be no delay between seeing ball hit and
hearing ball hit.
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Question:
You are at the ball park sitting in the bleachers in the
outfield (~325 ft from the batter). You see the bat hit
the ball. About how long will it take before you hear the
bat hit the ball?
c. About 0.3 seconds
How fast is that sound traveling?
Speed = distance/time = 325 ft/0.3 s = 1083 ft/s or 330 m/s
Speed of Sound in Air = 331 m/s at 0 degree C
343 m/s at 20 degree C
(Speed of Light = 3 X 108 m/s… much, much faster)
About 0.3 second means ~ 325 ft away from batter.
In 3 sec sound travels ~3,250 ft
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What causes the delay between seeing and hearing?
A sound wave travels or propagates through the air and
this takes a bit of time to get from the bat to your ear!
Useful visual picture of the stuff air is made of
… A bunch of nitrogen and oxygen molecules bouncing
around and colliding with each other and anything else
they come across.
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When you hear the crack of the bat with your ear,
what is it that your ear is detecting?
Clap is like when bat hits ball. When you clap, you push the
air causing a slight increase in the pressure of the air
followed by a slight decrease in pressure air.
This pressure fluctuation travels out in all directions as a
wave. As the pressure wave reaches your ear, you hear
sound.
AIR MOLECULES
Just after
clap
Slight decrease
More densely packed air molecule…
Slight increase in pressure
Later
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How does the wave travel?
a. The air molecules I push when I clap travel through air
between me and you, and then change the pressure at your
ear drum.
b. The air molecules I push when I clap will in turn push on the
air molecules near them and these will push on the air
molecules near them, etc, etc. And this is how the pressure
fluctuation travels between me and you.
AIR MOLECULES
Just after
clap
Later
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What happens in the wave?
Answer is b. The air molecules I push when I clap will in turn push on
the air molecules near them and these will push on the air
molecules near them, etc, etc. And this is how the pressure
fluctuation travels between me and you.
I push molecules and they push on molecules near them (Sound travels
like a domino effect).
When push comes to shove, all physics is local
Thomas (Tip) O’ Neil
Like a slinky….
AIR MOLECULES
Just after
clap
Later
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A sound wave is a longitudinal wave … like a
slinky:
Basically, this means that the motion of the air
molecules is towards and away from the source of the
sound.
This is in contrast to a typical wave, like the wave that forms
when I throw a stone in water. In this wave the water is
moving up and down, not towards and away from the source
of the wave.
Motion of air molecule – after sound
wave passes, molecule returns to
original position
Movement of air:
10-5 m for loudest tolerable sound
10-11 m for faintest of sound we can hear
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Creating Musical Tones
To create a pure sustained tone (like concert A), the
speaker pushes on the air at regular intervals and this
creates a series of pressure waves.
In speaker we vibrate cone:
Higher P
Lower P
All instruments work with same principle... push on air at
regular intervals.
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Look at sound:
Microphone detects changes in pressure.
Sound waves traveling out
Hit microphone,
It flexes, Creates
electrical signal
Higher P
Lower P
pressure
time
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Sound waves traveling out
Higher P
Lower P
If the speaker vibrating back and forth 200 times each second,
how much time passes between each time it produces a
maximum in pressure?
a. 0.2 seconds
b. 200 seconds
c. 0.005 seconds
d. 0.02 seconds
e. 0.05 seconds
Answer is c. 0.005 seconds.
Period = 1 second/ 200 cycles = 0.005 seconds/cycle
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What if we wanted to increase the pitch of the tone
produced by the speaker?
To get a higher pitch sound, we need to adjust the speaker
so that:
a. It vibrates back and forth more rapidly, taking a smaller
amount of time for each cycle
b. It vibrates back and forth at the same rate as before, but
the range of its back and forth motion is larger.
c. It vibrates back and forth more slowly, taking a longer
amount of time for each cycle
d. It vibrates back and forth at the same rate as before, but
the range of its back and forth motion is smaller.
Correct answer is a .… It vibrates back and forth more rapidly,
taking a smaller amount of time for each cycle … this is
adjusting the frequency
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The pitch of the sound is determined by the frequency of the
vibration:
Frequency:
The number of times per second that the speaker goes through
one complete pushing motion
or
The number of times per second that the pressure in my ear goes
through rise-fall cycle.
The frequency of Concert A is 440 Hz, or 440 cycles/second
FGAB | CDEFGAB | CDEFGAB |
Octave below Concert A
(220 Hz)
Logarithmic scale
middle C
(256 Hz)
Concert A
(440 Hz)
Octave above Concert A
(880 Hz)
Human Hearing: 20 Hz to 20,000 Hz
Hearing loss with age, rock concerts, shooting ….
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Wavelength
Higher P
Amplitude
Lower P
Question: If the speaker oscillates at 200 Hz (remember that is completing
one cycle in 0.005 seconds), what is the distance between the pressure
maxima (i.e. what is the wavelength)? Speed of sound =330 m/s
a. 0.6 m
b. 1.65 m
c. 66,000 m
d. 3.3 m
b. 1.65 m…. Sound will travel 330 m in 1 sec, will
oscillate 200 times in that 1 sec or once in 0.005
seconds. So length per oscillation is
Velocity * time = (330 m/s) x 0.005 seconds = 1.65 m
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Relationship between speed, period,
and wavelength!
• velocity = wavelength / period
• period = 1/frequency
• velocity = wavelength * frequency
• Period: T, Wavelength: λ, Frequency: f
v = λ/T = λf
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• velocity = wavelength / period
• period = 1/frequency
• velocity = wavelength * frequency
• Speed of sound is 330 m/s
• Subwoofer generates sound at 20 Hz, what is the
wavelength of that sound?
a) 16.5 m
b) 1.65 m
c) 1.65 Hz
d) 1.65 cm
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• velocity = wavelength / period
• period = 1/frequency
• velocity = wavelength * frequency
• Speed of sound is 330 m/s
• Subwoofer generates sound at 20 Hz, what is the
wavelength of that sound?
a) 16.5 m
b) 1.65 m
c) 1.65 Hz
d) 1.65 cm
Answer is a: wavelength = velocity/frequency
= 330 m/s / 20 Hz = 16.5 m
Longer than this room is tall!
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If the speaker vibrates back and forth twice as fast (so 40 times per second),
the period of the sound wave (i.e. the time between producing each peak
in pressure) is
a.
twice as long
b. half as long
c. doesn’t change
b. Half as long: Time between peaks = (1 sec)/(40 cycles)
= 0.025 seconds/cycle
The wavelength of the sound wave (i.e. the distance between
producing each peak in pressure) is
a. twice as long
b. half as long
c. doesn’t change
b. Half as long. Distance (wavelength) = velocity of sound x
time between peaks = 330 m/s x 0.025 seconds = 8.25 meters
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More on speed of sound through air:
all frequencies travel at same speed …
What would happen to orchestra music if frequencies traveled at
different speeds?
speed of sound in air is a fundamental property of the air
pressure and density
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Microphone
Higher P
Lower P
pressure
time
Question: If I increase the volume, what will happen to the signal from
the microphone?
a. The peaks will go up and the valleys will go down.
b. The peaks will get closer together.
c. Both a and b.
d. Nothing will happen
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Microphone
Higher P
Lower P
pressure
time
Question: If I increase the volume, what will happen to the signal from the
microphone?
Answer is a. The peaks will go up and the valleys will go down. When I
turn up the volume, the speaker cone moves further and piles up more
air molecules. High pressure is higher. Low pressure is lower. But on
average pressure is the same.
So a louder volume means a larger pressure difference between peak and
valley.
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Ear detects very, very small pressure changes:
Normal pressure of air (at sea) = 1 atmosphere
Minimum pressure change detectible by ear = 2 X 10-10 atmospheres
Maximum pressure change detectible by ear = 3 X 10-4 atmospheres
Regular conversation, pressure change ~ 3 X 10-7 atmospheres
So minimum change is 1 part in 5 billion, maximum is 1 part in 3600.
Volume scale is LOGARITHMIC
Measure volume in db (decibel): 1db=10*Log(P/Pref) (Log: base 10)
Most senses are logarithmic; gives greater range
3db ~ a factor of two (Log(2)=0.30)
Reference volume is arbitrary; usually set to nominal threshold of hearing
Whisper: 20db
Normal speech: 70db
Jet plane: 120db (pain threshold)
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Speakers
•
Two design problems with speakers:
1) Bass is weak, 2) treble is directional
Let’s consider (1)
demo
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Why is Bass Louder With Baffle?
A
B
C
D
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Bass Wavelengths too Long
At 50Hz, Wavelength=660 cm
.
Amplitude
Speaker~10cm
Entire speaker is at essentially the same pressure; no sound!
General property of ALL waves; we can only probe length scales that are
greater than the wavelength (same with radar, lasers, etc).
That’s why we need blue lasers to increase storage capacity of DVDs
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Woofer designs separate front and back waves
1) Bass Reflex
Enclosure has a port that channels the back
wave to the front. Port is tuned so that the
waves at the natural (resonance) frequency
of the driver emerge from the port in phase
with the front wave. Sometimes the port also
includes a “passive” cone.
This game has limits; the port can only be
tuned to one frequency. Bass reflex
speakers have less definition. Also, the
maximum enhancement one can expect is a
factor of two, i.e. only 3db
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Woofer designs separate front and back waves
2) Acoustic suspension
Enclosure is completely sealed
The air inside is used as a spring (hence
“suspension”) that complements the
suspension of the cone of the driver.
Acoustic suspension speakers are less
efficient than bass reflex, and must be
carefully tuned for the “air spring” to
work properly.
All higher-frequency drivers (midrage and
tweeter) are in an acoustic suspension
enclosure, even in bass reflex speakers.
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Woofer designs separate front and back waves
3) Transmission Line
Enclosure has a long tuned cavity that
channels the back wave out the back of
the speaker. The transmission line is
carefully tuned so that the resonant
wave emerges from the back of the
speaker with the correct phase. This
requires very long transmission lines, so
speakers are very large, even when the
tranmission line is folded (as it always is).
This is a rare design; it is very exacting and
the gain is not higher than bass reflex.
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String instruments: oscillators
• String is an oscillator
• Spring constant proportional to tension divided by
length, Tension/L
• Mass M is inertia
• Expect period proportional to sqrt(M/(Tension/L)) =
sqrt(ML/Tension)
• Actually, period = 2xsqrt(ML/Tension)
• Mass per unit length µ = M/L
• Period = 2x√ (ML/Tension) = 2x√ (µL2/Tension) so
Period = 2Lx√ (µ/Tension)
• proportional to length
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Can you answer?
• How does one get the frequency from the period?
• What is the speed of sound in air?
• How can I calculate the wavelength if I know the
period?
• What range of frequencies are audible?
• What is the fundamental period of a taught
string?
• Why do we use baffles in woofer design?
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