Chapter 16: Sound Waves and Hearing Hints&Answers-9/23/10
Sound waves are mechanical longitudinal waves that travel through the air or any other substance.
They are three-dimensional waves and allow us to communicate through speech. The speed of sound in
air at standard pressure and room temperature (1.013 x 105 Pa and 20ºC) is about 344 m/s, but it can
vary depending of air pressure and temperature. Sound can also travel through liquids and solids at
higher speeds.
1. Answer the following questions about sound waves.
a) How do we know that sound waves are longitudinal? Sound travels through gases, which cannot
sustain transverse waves. Another hint are the sources of sound, which are always back-and-forth
vibrators such as tuning forks, speakers, vocal chords, etc.
b) How do you know that (in a longitudinal wave) the oscillators in the compressions are
instantaneously moving in the direction of the waves and the ones in the rarefactions are moving in the
opposite direction? The argument is that for a compression to move to a spot further down the medium
the oscillators in the compression must be moving forward and the oscillators in the rarefaction must be
moving in the opposite direction “squeezing” the neutral region in between them so the compression
moves forward….
c) What is the purpose of the slide in a trombone or the valves on a trumpet or the frets in a guitar? All
these change the length of the pipe or string, which change the resonant wavelengths and frequencies
produced by the instrument, hence changing the note.
d) Explain why all harmonics are present in an organ pipe open at both ends, but only the odd
harmonics are present in a pipe closed at one end. The fundamental in the tube open at both ends is ½ a
wavelength, and the harmonics are all the multiples of that. The fundamental in the tube open-closed
tube is 1/4 of a wavelength, and the harmonics are only the odd multiples of that…
e) Compare the frequency, wavelength, and speed of the standing wave in a string of a violin and the
wave that travels to your ear. The frequency in the string and the sound produced is the same since the
source is the string. The speed in the string depends on the tension and density of the string and is
generally faster than the speed of sound in air that depends on the properties of the air. The
wavelengths would be proportional to the wave speed. In addition, the string waves are transverse and
the sound waves are longitudinal.
f) Explain how the distance to a lightning bolt may be determined by counting the seconds between the
visible flash and the heard sound of the thunder. Why can we neglect the speed of light? You can
neglect the speed of light because it is so much faster than sound, and w cane assume that we see the
flash nearly at the instant it occurs. Counting the seconds from flash to thunder allows us to use the
speed of sound to estimate the distance to the location of the lightning.
2. A stone is dropped into a deep canyon and is heard to strike the bottom 10 s after release. Remember
g=9.8 m/s2 and vsound=344 m/s. This information can be used to determine the depth of the canyon.
a) As a first approximation you ignore the time it takes the sound to travel from the bottom of the
canyon. What is this approximation to the depth of the canyon?
a) Using d=gt2/2 and assuming that 10 s is the falling timed~490 m.
b) Use your approximation in (a) to determine the time it took the sound to travel from the bottom to
your ear. Was it justifiable to ignore that time?
b) The time for sound to travel 490m back to the listener is tup =d/v=490/344=1.4 s which seems too
large to neglect (~14% error).
c) You can get a better approximation to the depth of the canyon by subtracting the time found in (b)
from the original 10 s and doing (a) again. What % error would result from ignoring the speed of sound
the first time? (You can get an exact answer by not neglecting the sound in the place and solving a
quadratic equation.
c) A better approximation would be to subtract 1.4 s from 10 s (tdown =8.6 s) and use that as the “falling
time”d~362 m. But the time correction (1.4 s) was too large so we can do this again to get a better
time correction (tup =362/344 = 1.05 s). The new approximation: tdown=8.95 s d~392 m…After a few
of these iterations you zero in on the correct answer of 386 m. Or you can get the correct answer by
solving the quadratic equation: 10 = tdown+ tup = (2d/g)1/2 +d/vsound)…
3. There are a number of formulas that predict the speed of waves in different media. Review the ones
discussed in class.
a) The Bulk modulus is equal to P where is a constant equal to 7/5 for air and P is the air pressure.
Determine the speed of sound at 1 atm and 0ºC when the density of air equals 1.29 kg/m3.
a) 1atm= 101 kPa…then just plug into formula: v=(B/ = v=(P /= 331 m/s
b) By proper excitation it is possible to produce both longitudinal and transverse waves in a long metal
rod. A particular metal rod is 150 cm long and has a radius of 0.20 cm and a mass of 50 g. The Young’s
modulus for the material is 6.8 x 1010 N/m2. What must be the tension in the rod if the ratio of the speed
of longitudinal waves to the speed of the transverse waves is 8 to 1?
b) vlong =(Y/ ; vtrans =(FT/µ ; and µ=A. We want the ratio (vlong /vtrans)=8/1…after substituting
and solving for FT, and making sure units were compatible, I got ~13 kN.
4. A “wave splitter” consists of two tubes, each 1.75 m long. The top tube contains air at 0ºC (vsound=
331 m/s) and the lower tube contains air at 200ºC (vsound=435 m/s). Sound signals produced by a source
travel through the tubes and the two signals meet again at a receiver at the other end of the tubes.
a) What is the lowest frequency that will produce an intensity minimum at the receiver?
a) The no. of waves inside each tube are: ntop=1.75f/331 and nbot=1.75f/435. The difference has to equal
½ for the first minimum. So, (1.75f/331 -1.75f/435) =0.5, and solving for “f” I got ~396 Hz.
In this problem you have to compare “# of waves” rather than the tube lengths (the two tubes have
equal lengths here) because the wavelengths are different in each tube. What matters for interference is
the phase difference and  is proportional to the difference in the no. of wavelengths  /2π=[(#waves
in tube1) - (#waves in tube2)], and for a min  =±π. There are other possible frequencies as well, since
destructive interference can also result from  =±3π, 5π, etc…
Another way of looking at this problem is to compare the phase difference due to the different times
taken to travel the tube. Since  is proportional to ∆t /2π=∆t/T=f∆t, and ∆t=(L/v1)-(L/v2). You end
up with the same relationship.
b) How much would you have to change the length of one tube to turn that minimum into a maximum?
b) There are various ways to do this: make the top or bottom tube ½ wavelength shorter or longer. For
example, in the top tube  =0.42m so making that tube that much shorter or longer should do it.
5. Two loudspeakers are placed on a wall 3 m apart. A listener stands directly in front of one of the
speakers 4 m from the wall. The speakers are driven by a single oscillator at a frequency of 275 Hz.
Assume the speed of sound is 344 m/s.
a) What is the phase difference between the two waves when they reach the observer?
a) The wavelength =1.25 m, and =(∆r/)2π=1.6π. This is between a max and a minimum.
b) What is the frequency closest to 275 Hz to which the oscillator may be adjusted such that the
observer will hear minimal sound?
b) Now you want  =π, and solving for the wavelength and then the frequency I got 172 Hz.
6A. a) Broken lines are antinodal lines
and solid lines are nodal lines.
b) In the first case the separation is 2
wavelengths and there are 2 nodal
lines and 1 antinodal line in each
quadrant, the axes are both antinodes.
In the second case the separation is 3.5
wavelenghts and there are 3 nodal lines
and 3 antinodal lines; the x-axis is an
antinode and the y-axis is a node.
6. Two speakers are driven by the same oscillator of frequency 200 Hz. They are located on the vertical
pole a distance 4 m from each other. A man walks directly toward the lower speakers in a direction
perpendicular to the pole as shown. Take the speed of sound to be 340 m/s and ignore any sound
reflections from the floor.
a) How many times will he hear a minimum in sound intensity? Hint: As he
walks toward a speaker he crosses the nodal lines in that quadrant.
L
4m
The no. of nodal lines in one quadrant equals the no. of full waves that fit in
between the speakers. Here #nodal lines=4/=4f/v=2.4 so he will hear 2 minima.
b) How far is he from the wall at these moments?
b) Call the distance from the speaker “L”. Then the path difference is ∆r=[(L2+16)1/2 –L]and for a
minima ∆r=nodd /2.
Solving this equation for L gives two possible results for nodd =1 and 3. My answers: L1=8.99 m and
L3=1.86 m
7. Clearly distinguish between the pressure amplitude and the displacement amplitude of a sound wave.
Review their relationship to each other and to the wave functions. Review your notes.
a) An experimenter wishes to generate a sound wave in air with a displacement amplitude equal to 5.5
µm. The pressure amplitude is to be limited to 0.84 Pa. What is the minimum wavelength that the sound
can have?
a) The density and wave speed at room temperature are 1.2 kg/m3 and 344 m/s. Substitute into
∆pmax=vS to find  then f, then . I got 5.8 m. Longer wavelengths will generate less pressure.
b) Write an expression that describes the pressure variation as a function of position and time for a onedimensional sinusoidal sound wave described in (a).
b) ∆p(x,t)= ∆pmax sin (kx –t); k=2π/5.8 rad/m and =118π rad/s.
c) Write the function that describes the displacement wave corresponding to this pressure wave.
c) As discussed in class, pressure and position are 90º out of phase: s (x,t)=S cos(kx –t)
8. For sound, the intensity is a very useful concept (I=Power/Area). Assume that the intensity of a sound
wave at a fixed distance from a speaker vibrating at 1 kHz is 0.60 W/m2. Review the formulas for the
power and intensity of a sound wave and answer the following questions. We assume v and  don’t
change.
a) Determine the intensity at the receiver if the frequency of the source is increased to 2.5 kHz while
keeping the displacement amplitude constant. What happens to the pressure amplitude in this case? Has
the power output of the source of sound changed?
a) Since S is not changing and I=vS2)/2 =(∆pmax2)/2v, I increases by 2.526.25(0.6)= 3.75W/m2.
The pressure amplitude increases by 2.5, since ∆pmax=vS.
Since the intensity is increasing at the same distance from the source, the power output (P=IA) of the
source must have increased by the same factor 2.52.
b) Determine the intensity at the receiver if the frequency of the source is reduced to 0.50 kHz while the
displacement amplitude is doubled. What happens to the pressure amplitude in this case? Has the power
output of the source changed? No net change in I, power, or ∆pmax, since the change in frequency and
displacement amplitude cancel each other out.
c) Assuming that the displacement amplitude of the source doesn’t change, determine the intensity at
the receiver if the frequency of the source is increased to 2.0 kHz and the receiver is moved twice as far
away from the source (assume it is a point source). What happens to the pressure amplitude in this case?
Has the power output of the source changed?
The intensity and pressure amplitude are the same because the frequency doubles but the displacement
amplitude at the receiver is halved (due to the inverse squared law distance factor in the area). The
power output of the source must have increased by 4.
d) Assuming that the power output of the source doesn’t change, determine the intensity at the receiver
if the frequency of the source is increased to 2.0 kHz and the receiver is moved twice as far away from
the source (assume it is a point source). What happens to the pressure and displacement amplitudes in
this case?
In this case the intensity at twice the distance is ¼ because of the inverse-squared area factor,
regardless of the frequency change. The amplitudes change to compensate for the frequency change.
The pressure amplitude is proportional to √I, regardless of frequency [∆pmax= √(2Iv)] so ∆pmax
decreases by ½ .
The displacement amplitude is proportional to (∆pmax /)[Smax= ∆pmax/v] so Smax decreases by ¼ .
9. When adding sound levels you have to consider that what actually “adds up” are the intensies
(power/area) of the sources.
a) Suppose two sources produce sound levels of 75 dB and 80 dB separately at a particular location.
What is their combined intensity in W/m2?
a) Since ß= 10log(I/Io)I= Io 10 ß/10…then Ia= Io(107.5); Ib= Io(108); and Iab= Io(1.32 x108) where
Io=(10-12).
b) What is their combined sound level? ßab= 10log(1.32 x108)=81 dB.
c) Prove that, in general, when adding two sources of sound levels (ßa and ßb), the resultant sound level
is (ßab)= 10log[10ßa/10 + 10ßb/10]. Follow the method in (a) and (b)…
10. Because the decibel scale is defined as a log function, when comparing dB levels, a difference is
more logical comparison than a ratio. For example, suppose you want to compare the sound level of
one source of sound to “N” sources of the same sound. It is easier to determine (ßN – ß1), than (ßN /ß1).
a) Derive a formula for the differential comparison (ßN – ß1).
a) (ßN – ß1)= 10log(NI/Io) - 10log(I/Io)= 10log(NI/I)= 10log(N)
b) As a numerical example, imagine that in the afternoon the sound level of busy freeway is 80 dB with
100 cars passing a given point every minute. Late night the traffic flow is only five cars per minute.
What is the late night sound level?
b) (ß100 – ß5)= 10log(100/5)=13 ß5=80 – 13=67dB
11. A point source of sound would produce waves that spread out spherically from the center. Since
intensity is the power/area (I=P/A) and the area of a sphere equals 4πr2, the intensity in this case is
“inverse-squared” related to the distance from the source “r”. a) Show that the difference in decibel
levels, 2 and 1, from a point sound source is related to the ratio of the distances, r2 and r1, from the
receivers by the formula: 2 - 1= 20 log(r1/ r2).
2 - 1= 10log(P/4πr22Io) - 10log(P/4πr12Io)= 10log(r12/r22) =20 log(r1/ r2).
b) If you are close to a long vibrating string, the source of the sound is more “line-like” than “pointlike” and the sound initially spreads out through cylindrical (rather than spherical) cross-sectional areas.
Show that in this case, the difference in decibel levels at distances r2 and r1 from a line sound source is
2 - 1= 10 log(r1/ r2). Ignore the relatively small areas of the top and bottom of the cylinder.
b) Here the areas are 2πrL, so 2 - 1= 10log(P/L2πr2Io) - 10log(P/L2πr1Io)= 10log(r1/r2) =10 log(r1/ r2).
12. A fireworks rocket explodes at a height of 100 m above ground. An observer on the ground directly
under the explosion experiences an average sound intensity of 0.07 W/m2 for 0.2 s.
a) What is the total sound energy of the explosion?
a) We assume that the explosion acts like a point source. Intensity=Power/Area=(Energy/time)/Area,
and for a point source Area=4πr2. Solving for Energy=IAt=(0.07)4π(100)2(.2)=1.76 kJoules
b) What is the sound level in decibels heard by the observer?
b) Since ß= 10log(I/Io)=10log(0.07/10-12)=108 dB
c) What is the sound level in decibels heard by another observer 50 m from the first observer?
b) The second observer is 112 m from the explosion. Given the inverse-square law, ß’=
10log(I’/Io)=10log[(0.07x1012)(100/112)2]=107 dB
13.The pressure changes generated by spherical wave radiating from a point source is described by the
following formula:
p(r,t)= (25/r) sin (1.25r – 1870t), where p is the gauge pressure ∆p in pascals,
r is in meters, and t in seconds.
a) What is the maximum pressure amplitude 4 m from the source? Why does the amplitude decrease as
1/r instead of as 1/r2? pmax=25/4=6.25 Pa. The amplitude decreases as 1/r because the power (and
therefore the intensity) depend on the amplitude squared, so when the intensity decreases by 1/r2 the
amplitude must decrease by 1/r to reflect the decrease in intensity.
b) Determine the speed of the wave. If you have access to a table of wave speeds in different materials
you can determine the material the wave is in.
b) From the given equation k=1.25 and =1870v=/k=1496 m/s. This is close to the speed of sound
in water so if we assume water is the medium here.
c) Find the intensity of the wave in dB at a distance 4 m from the source.
c) Recall that I=(pmax2)/2v and, assuming the medium is water, =1000 kg/m3. Plugging in:
I=(6.25)2/2(1000)1496=1.3 x 10-5 W/m2 and ß= 10log(1.3 x 107)=71 dB
d) Find the instantaneous pressure 5 m from the source at 0.08 s.
d) More plug-ins: p=(25/5) sin(1.25x5 -1870x0.08)=5x(0.918)=4.59 Pa
14. Two small speakers emit sound waves of different frequencies. Speaker A has an output of 1.0 mW
and speaker B has an output of 1.5 mW. Determine the sound intensity level (in dB) at point C is 5 m
from A and 4 m from B. Note: if the frequencies are different, superposition cannot generate a fixed
interference pattern in space and there are no max or min spots.
a) Determine the sound intensity level (in dB) at point C if: (i) only speaker A emits sound; (ii) both
speakers emit sound.
a) Keep in mind that because the frequencies are different a fixed interference pattern in space cannot
emerge and the sources behave simply as two independent sources of sound. (i) Ia=(0.001/4π25)= 3.18
x10-6 W/m2  ßa= 10log(3.18 x10-6/10-12)=65 dB.(ii) Ib=(0.0015/4π16)= 7.46 x10-6 W/m2  ßab=
10log(10.64 x10-6/10-12)=70 dB.
b) If both speakers had the same frequency, interference could make point C a local maximum or
minimum, or something else. How would this affect the answers in (a)?
b) If the point were a minima the intensity of the sound would be lower, and if it were a maxima it would
be higher. Not knowing the frequency here it’s impossible to get a numerical answer. But the method
would involve finding the actual amplitude at C and to be able to find the intensity at C.
c) For a numerical version of (b), assume that the speed of sound is 340 m/s and the sources have the
same frequency of 170 Hz. Verify that the point in question is a minimum, and determine the decibel
sound level.
c) The wavelength here is 2 m, and since the path difference from the sources is 1 m, the point is clearly
a minimum, which means the amplitudes from the two sources subtract. You can determine the
displacement amplitudes to be 0.115 µm and 0.179 µm. The net amplitude is 0.064 µm and sound level
is 60 dB.
15. The questions and problems below deal with the Doppler effect and related phenomena. You should
be able to explain the following:
a) Explain why the Doppler-shifted frequency heard by the listener when he’s moving toward the source
is NOT the same as when the source is moving toward the listener. Why is this NOT a violation of the
Law of Inertia? Check your class notes. In one case the wavelength is changed, in the other case, the
apparent speed of the wave is changed.
b) Highways often display signs that say “Speed checked by radar”. Explain how police use the Doppler
effect with microwaves to determine the speed of an automobile. The measured frequency reflected
from a moving source is related to the speed of the source (check the formulas), so the speed of the
moving car can be determined from the echo of the radar that is aimed at it by the police.
c) Suppose that an observer and a source of sound are both at rest and a strong wind blows toward the
observer. Describe the effect of the wind (if any) on: (i) the observed wavelength, (ii) the observed
frequency, and (iii) the wave velocity. If you check the formulas, the wind would increase the
wavelength and also make the apparent speed of the wave faster by the same factor. Therefore the
frequency heard by the listener will be the same as that of the source and a Doppler shift would not take
place.
d) A binary star system consists of two stars revolving about each other. If we observe the light reaching
us from one of these stars as it makes one complete revolution about the other, what does the Doppler
effect predict will happen to the observed light? You should know that the frequency of light determines
its color.
d) The frequency observed would be lower when the star is moving away and higher when the star is
moving toward the observer. Since frequency determines the color of light, each star’s “color” would
change as it revolves around the other…
Note: When doing Doppler effect problems it is easy to get the wrong sign. One of the “checks” it to
realize that, whichever way the vehicles are moving, if they are getting closer the heard frequency will
be higher than the source frequency, and if they are getting farther apart the heard frequency will be
lower than the source frequency. If your answer violates this “rule” you’ve made a mistake with the
signs somewhere!
16A. Review the Doppler effect for the basic “moving source” and “moving listener” cases.
a) Explain why the difference in the frequency heard by the listener in these two cases is not a
violation of Galilean relativity. The two cases are not equivalent because the wave travels in the
medium…
b) How would changing frames of reference affect the Doppler formula? It would change the
speeds of the source and listener…
c) Compare the Doppler examples of a sound reflecting off a reflector at rest (echo example) and of
sound reflecting of a reflector in motion (radar example). The net effect is the same…
d) How do we handle the case of a moving medium, as when the air itself is moving with some
wind velocity? You place yourself in the frame of reference of the wind and determine the speed
of the source and listener in this reference frame. Then use the Doppler formula.
16. A fire truck moving to the right at 40 m/s sounds its horn (f=500 Hz) at the two vehicles (a car and a
van) ahead of it. The car is moving to the right at 30 m/s while the wan is at rest.
Assume vsound=344 m/s. When using Doppler formula make sure
you understand the ± in the terms…
a) What frequency is heard by the passengers in the car?
a) Using Doppler formula: f’=500[(344-30)/(344-40)]=516 Hz
b) What frequency is heard by the passengers in the van?
b) f’=500[(344-0)/(344-40)]=566 Hz
c) When the fire truck is 200 m from the car and 250 from the van, the passengers in the car hear a sound
intensity level of 90 dB. At that time, what is the intensity level heard by the passengers in the van?
c) Assuming that the truck’s horn acts like a point source, we can use the sound level equation: (2 - 1)=
20 log(r1/ r2) ( - 1)= 20 log(250/200)=1.93  1~ 88 dB
17. A train is moving parallel to a highway at 20 m/s. A car is traveling in the opposite direction as the
train at 40 m/s. The car horn sounds at 510 Hz and the train whistle sounds at 320 Hz. Assume the train
and car are in line.
a) When the car is behind the train what frequency does a car passenger hear for the train whistle?
a) Here the train and car are moving in opposite directions and getting further apart, so the wavelength
reaching the car is longer and the speed of the wave appears slower: f’=320[(344-40)/(344+20)]=300
Hz
b) What size wavelength reaches the car?
b) The wavelength is longer than if train weren’t moving: ’= +vsourceT = (vwave+vsource)/f=
364/320=1.14 m.
c) When the car is in front of the train what frequency does a train passenger hear for the car horn?
What size wavelength reaches the train?
c) Here the train and car are moving in opposite directions but getting closer. The wavelength reaching
the train is shorter and the speed of the wave appears faster: f’=510[(344+20)/(344-40)]=611 Hz. The
wavelength reaching the train is shorter: ’= -vsourceT =( vwave -vsource)/f=304/510=0.596 m.
18. Two ships are moving along a line due east. The trailing vessel has a speed of 64 km/h, and the
leading ship has a speed of 45 km/h relative to a land-based observation point. The two ships are in a
region of the ocean where the current is moving uniformly due west at 10 km/hr. The trailing ship
transmits a sonar signal at a frequency of 1200 Hz.
a) What frequency is monitored by the leading ship? The speed of sound in ocean water is 1520 m/s
(5472 km/hr).
a) You need to match speed units. It is easier in this problem to change the speed of sound to km/hr:
1520 m/s =5472 km/hr. The other important detail here is that the speed of the current effectively
reduces the speed of sound by 10 km/hr (v’sound =5462 km/hr) as it travels from the trailing to the
leading ship. You can also put yourself in the current frame of reference and the speed of the vessel
would increase in that frame by 10 km/hr. Either way, applying the Doppler formula gives:
f’=1200[(5472-10-45)/(5472-10-64)]=1204 Hz.
b) Redo the problem assuming that the trailing ship is actually moving west.
b) Here f’=1200[(5472-10-45)/(5472-10+64)]=1176 Hz
19. A skydiver carries a tone generator to determine her speed when falling. A friend on the ground
directly below has a receiver. While the skydiver is falling at a constant terminal speed, her tone
generator emits a steady tone of 1800 Hz
a) If the friend on the ground receives waves of frequency 2150 Hz, what is skydiver’s speed?
a) Applying Doppler formula: f’=2150=1800[(344)/(344- vdiver)]  vdiver =56 m/s
b) If the skydiver were also carrying a receiver sensitive enough to detect waves reflected from the
ground, what frequency would she receive?
b) This is an “echo problem” and the frequency reflected is 2150 Hz: fecho = freflected[(344+56)/(344)]=
2150[(344+56)/(344)]=2500 Hz
c) What size wavelength reaches the receiver on the ground? What size wavelength reaches the skydiver
after reflection from the ground?
c) Wavelength is shorter: ’= -vsourceT = (vwave- vsource)/f=[(344-56)/(1800)]=0.16 m. The reflected
wave has the same wavelength since the reflector (the ground) is not moving.
20. Bats navigate using sound echoes, a skill called echolocation. Suppose a bat moving at 5 m/s is
chasing a flying insect. The bat emits a 40 kHz shirp and receives back an echo of 40.4 kHz.
a) At what speed is the insect moving toward or away from the bat? Is it possible to tell whether the
insect is moving toward or away from the bat?
a) The solution is unique and you will find that (whatever assumption you make initially) the insect is
moving in the same direction as the bat but slower (at 3.3 m/s). This is the most intricate of the Doppler
problems because it involves an echo from a moving “listener-turned-into-source” (the insect in this
case). Applying the formula: fecho=40.4 = freflected[(344+5)/(344+vinsect)] = fbat[(344-vinsect)/(3445)][(344+5)/(344+vinsect)] =40[(344-vinsect)/(344-5)][(344+5)/(344+vinsect)] vinsect=3.3 m/s
b) What is the wavelength of the sound generated by the bat? The bat’s motion shortens the wavelength
bat = (344-5)/40 = 8.48 mm
c) What is the wavelength of the sound reflected by the insect? The insect’s motion lengthens the
wavelength, but it is also the wavelength that the bat receives on reflectioninsect =(344+5)/40.4
=8.64 mm
21. A supersonic jet traveling at Mack 3 (3x the speed of sound) at an altitude of 20 km is directly
overhead at time t=0.
a) How long will it be before a person on the ground encounters the shock wave?
a) According to the shock-wave formula the angle ø=arcsin (1/3)=19.5º. Using the geometry of the
triangle when the bow hits the person you can find the distance “x” the plane traveled from when it was
directly over the person tanø=h/x x=h/tanø x=20/tan19.5º = 56 km. Finally you can use the
speed of the plane to find the time: t=x/v=56,000/(344)3=55 sec
b) Where will the plane be when it is finally heard?
b) This the “x” found above: 56 km.
ø
t=0
ø
h
Observer hears
boom
22. An interesting example of “shock waves” is the Cerenkov effect, which occurs when high-energy,
charged particles move through a transparent medium with a speed greater than the speed of light in that
medium. Even though particles cannot travel faster than light in a vacuum, light “travels” slower than
“c” in transparent media such as water or glass. The Cerenkov effect can be observed in the vicinity of
the core of a swimming-pool reactor due to high-speed electrons moving through the water. We will
learn more about light waves in the optics unit of this course.
a) Calculate the speed of electrons in the water, given that the speed of light in water is 2.25 x 108 m/s
and the angle of the “shock bow” is 53º. sin 53º= 2.25 x 108 /velectrons velectrons =2.81 x 108 m/s.
b) What is the ratio of the speed of light in a vacuum (c=3 x 108 m/s) to the speed of light in water. This
ratio is called the index of refraction of water. Answer: 1.33
23. While attempting to tune the note C at 523 Hz, a piano tuner hears 2 beats/s between the reference
oscillator and the string.
a) What are the possible frequencies of the out-of-tune string? Either 525 Hz or 521 Hz.
b) When the string is slightly tightened, the beat frequency changes to 3 beats/s. What is the frequency
of the string?
b) Tightening the string increases the speed of the wave in the string, which raises the frequency it
produces. This takes the string further out of tune (3 beats), so the string’s original frequency must have
been the higher of the two possibilities in (a) 525 Hz
c) By what percentage should the tension in the string be changed to bring it into tune?
c) You can show that f 2 is proportional to F...so the tension should be reduced by the factor
(523/525)2=0.992This amounts to about a 0.8% change.