College Physics 140 Chapter 12: Sound
We will be investigating the sound waves!
Chapter 12: Sound
• Sound Waves
• The Speed of Sound
• Amplitude & Intensity of Sound Waves
• Standing Sound Waves
• Beats
• The Doppler Effect
• Echolocation
Sound Waves
Sound waves are longitudinal. They can be represented by either variations in pressure (gauge pressure) or by displacements of an air element.
The middle of a compression (rarefaction) corresponds to a pressure maximum (minimum). The Speed of Sound Waves
The speed of sound in different materials can be determined as follows:
In ideal gases
In thin solid rods
v=
v=
B
ρ
Y
ρ
B is the bulk modulus of the fluid and ρ its density.
Y is the Young’s modulus of the solid and ρ its density.
In fluids
v = v0
T
T0
Here v0 is the speed at a temperature T0 (in kelvin) and v is the speed at some other temperature T (also in kelvin).
For air, a useful approximation to the above expression is
v = (331 + 0.606TC ) m/s
where Tc is the air temperature in °C.
Materials that have a high restoring force (stiffer) will have a higher sound speed.
Materials that are denser (more inertia) will have a lower sound speed.
Example (text problem 12.6): A copper alloy has a Young’s Modulus of 1.1×1011 Pa and a density of 8.92×103 kg/m3. What is the speed of sound in a thin rod made of this alloy?
Y
1.1×1011 Pa
v=
=
= 3500 m/s
3
3
ρ
8.9 ×10 kg/m
The speed of sound in this alloy is slightly less than the value quoted for copper (3560 m/s) in table 12.1.
Example (text problem 12.1): Bats emit ultrasonic sound waves with a frequency as high as 1.0×105 Hz. What is the wavelength of such a wave in air of temperature 15.0 °C? v = (331+ 0.606TC ) m/s
The speed of sound in air of this temperature is 340 m/s.
v
340 m/s
−3
λ= =
=
3.4
×10
m
5
f 1.0 ×10 Hz
Example (text problem 12.4): A lightning flash is seen in the sky and 8.2 seconds later the boom of thunder is heard. The temperature of the air is 12.0 °C.
(a) What is the speed of sound in air at that temperature?
v = (331+ 0.606TC ) m/s
The speed of sound in air of this temperature is 338 m/s.
(b) How far away is the lightning strike?
d = vt = (338 m/s) (8.2 s) = 2800 m = 2.8 km
The speed of light is 3.00×105 km/s. How long does it take the light signal to reach the observer?
d
2.8 km
−6
t= =
= 9.3 ×10 sec
5
v 3.0 ×10 km/s
Amplitude & Intensity of Sound Waves
For sound waves:
2
0
I∝p
2
0
I ∝s
p0 is the pressure amplitude and
s0 is the displacement amplitude.
The intensity of sound waves also follow an inverse square law. Loudness of a sound is measured by the logarithm of the intensity.
The threshold of hearing is at an intensity of 10-­‐‑12 W/m2.
I
β = (10dB )log
I0
Sound intensity level is defined by
dB are decibels
Example (text problem 12.12): The sound level 25 m from a loudspeaker is 71 dB. What is the rate at which sound energy is being produced by the loudspeaker, assuming it to be an isotropic source?
Given:
I
β = (10dB) log = 71 dB
I0
Solve for I, the intensity of a sound wave:
log
I
= 7.1
I0
I
= 10 7.1
I0
I = I 010 7.1 = (10 −12 W/m 2 ) (10 7.1 ) = 1.3×10 −5 W/m 2
P
2
4
π
r
The intensity of an isotropic source is defined by:
P = I 4πr 2
I=
2
= (1.3 ×10 −5 W/m 2 )4π (25 m )
= 0.10 Watts
Example: Two sounds have levels of 80 dB and 90 dB. What is the difference in the sound intensities?
I
β1 = (10dB) log = 80 dB
I0
Subtracting:
I
β2 = (10dB) log = 90 dB
I0
"
I2
I1 %
β2 − β1 = 10 dB = 10 dB$ log − log '
I0
I0 &
#
"
I2 %
10 dB = 10 dB$ log '
I1 &
#
" I2 %
log $ ' = 1
# I1 &
I2
and I2 = 10I1
= 101
I1
Standing Sound Waves
Consider a pipe open at both ends:
The ends of the pipe are open to the atmosphere. The open ends must be pressure nodes (and displacement antinodes).
The distance between two adjacent antinodes is ½λ. Each pair of antinodes must have a node in between. The fundamental mode (it has the fewest number of antinodes) will have a wavelength of 2L.
The next standing wave pagern to satisfy the conditions at the ends of the pipe will have one more node and one more antinode than the previous standing wave. Its wavelength will be L. The general result for standing waves in a tube open at both ends is 2L
λn =
n
v
where n = 1, 2, 3,…
nv
fn =
=
= nf1
λn 2 L
f1 is the fundamental frequency.
Now consider a pipe open at one end and closed at the other.
As before, the end of the pipe open to the atmosphere must be a pressure node (and a displacement antinode).
The closed end of the pipe must be a displacement node (and a pressure antinode).
One end of the pipe is a pressure node, the other a pressure antinode. The distance between a consecutive node and antinode is one-­‐‑quarter of a wavelength.
Here, the fundamental mode will have a wavelength of 4L.
The next standing wave to satisfy the conditions at the ends of the pipe will have one more node and one more antinode than the previous standing wave. Its wavelength will be (4/3)L. The general result for standing waves in a tube open at one end and closed at the other is 4L
λn =
n
v
where n = 1, 3, 5,… (odd values only!!)
nv
fn =
=
= nf1
λn 4 L
f1 is the fundamental frequency.
Example (text problem 12.22): An organ pipe that is open at both ends has a fundamental frequency of 382 Hz at 0.0 °C. What is the fundamental frequency for this pipe at 20.0 °C?
At Tc = 0.0 °C, the speed of sound is 331 m/s.
At Tc = 20.0 °C, the speed of sound is 343 m/s.
v
v
=
1 =
λ1 2L
The fundamental frequency is f
The ratio of the fundamental frequencies at the two temperatures is: How long is this organ pipe?
Using either set of v and f1.
v
2L
v
L=
= 0.43 m
2 f1
f1 =
v20
f1,20
v20
2L
=
=
= 1.04
f1,0 v0
v0
2L
f1,20 = 1.04 f1,0 = 396 Hz
Beats
When two waves with nearly the same frequency are superimposed, the result is a pulsation called beats.
Two waves of different frequency
Superposition of the above waves
The beat frequency is
Δf = f1 − f 2
If the beat frequency exceeds about 15 Hz, the ear will perceive two different tones instead of beats.
The Doppler Effect
When a moving object emits a sound, the wave crests appear bunched up in front of the object and appear to be more spread out behind the object. This change in wave crest spacing is heard as a change in frequency.
The results will be similar when the observer is in motion and the sound source is stationary and also when both the sound source and observer are in motion.
The Doppler Effect formula
" vo %
$ 1− '
v 'f
fo = $
s
v
$ 1− s '
#
v&
fo is the observed frequency.
fs is the frequency emiged by the source.
vo is the observer’s velocity.
vs is the source’s velocity.
v is the speed of sound.
Note: take vs and vo to be positive when they move in the direction of wave propagation and negative when they are opposite to the direction of wave propagation. Example (text problem 12.39): A source of sound waves of frequency 1.0 kHz is stationary. An observer is traveling at 0.5 times the speed of sound. (a) What is the observed frequency if the observer moves toward the source?
fo is unknown; fs= 1.0 kHz; vo = -­‐‑0.5v; vs = 0; and v is the speed of sound.
" vo %
" −0.5v %
$ 1− '
$ 1−
'
v
v
fo = $
' fs = $
' fs = 1.5 f = 1.5 kHz
$ 1− vs '
$ 1− 0 '
#
#
v&
v &
(b) Repeat, but with the observer moving in the other direction.
fo is unknown; fs = 1.0 kHz; vo = +0.5v; vs = 0; and v is the speed of sound.
" vo %
" +0.5v %
$ 1− '
$ 1−
'
v
v
fo = $
' fs = $
' fs = 0.5 f = 0.5 kHz
$ 1− vs '
$ 1− 0 '
#
#
v&
v &
Echolocation
Sound waves can be sent out from a transmiger of some sort; they will reflect off any objects they encounter and can be received back at their source. The time interval between emission and reception can be used to build up a picture of the scene.
Echolocation
Sound waves can be sent out from a transmiger of some sort; they will reflect off any objects they encounter and can be received back at their source. The time interval between emission and reception can be used to build up a picture of the scene.
Example (text problem 12.47): A boat is using sonar to detect the bogom of a freshwater lake. If the echo from a sonar signal is heard 0.540 s after it is emiged, how deep is the lake? Assume the lake’s temperature is uniform and at 25 °C.
The signal travels two times the depth of the lake so the one-­‐‑way travel time is 0.270 s. From table 12.1, the speed of sound in freshwater is 1493 m/s.
depth = vΔt
= (1493 m/s) ( 0.270 s) = 403 m
Example (text problem 12.49): A bat emits chirping sounds of frequency 82.0 kHz while hunting for moths to eat. If the bat is flying toward a moth at a speed of 4.40 m/s and the moth is flying away from the bat at 1.20 m/s, what is the frequency of the wave reflected from the moth as observed by the bat? Assume T = 10.0 °C.
The speed of sound in air of C
this temperature is 337 m/s.
v = (331+ 0.606T
) m/s
The flying bat emits sound of f = 82.0 kHz that is received by a moving moth. The frequency observed by the moth is: ⎛ vo
⎜ 1 −
v
f o = ⎜
⎜ 1 − vs
⎜
v
⎝
+ 1.2 m/s ⎞
⎞
⎛
⎟
⎜ 1 −
⎟
337 m/s ⎟(82.0 kHz ) = 82.8 kHz
⎟ f s = ⎜
⎟
⎜ 1 − + 4.4 m/s ⎟
⎜
⎟
⎟
337 m/s ⎠
⎝
⎠
⎛ vo
⎜ 1 −
v
f o = ⎜
⎜ 1 − vs
⎜
v
⎝
⎞
⎛ − 4.4 m/s ⎞
⎟
⎜ 1 −
⎟
337 m/s ⎟(82.8 kHz ) = 83.6 kHz
⎟ f s = ⎜
⎟
⎜ 1 − − 1.2 m/s ⎟
⎜
⎟
⎟
337 m/s ⎠
⎝
⎠
Some of the sound received by the moth will be reflected back toward the bat. The moth becomes the sound source (f = 82.8 kHz) and the bat is now the observer.