Physics 1240
Homework 2 solutions
Brief solutions to homework set #2. Your numbers will be different, but the idea is the same.
I’m just using sample numbers here.
1. Decide whether each of the following statements are true or false.
(a) In this simulation, a darker shade of gray indicates a decrease in pressure compared to
the undisturbed air pressure.
True. I determined this by looking at the last panel of the simulation, where the air
pressure can be reduced to zero by removing all the air from the jar. As the air pressure
decreased, the shading inside the jar became a darker gray.
(b) If the speaker produces larger fluctuations in pressure, the volume of the tone heard by
the listener increases.
True. I determined this by increasing the amplitude with the slider control with the
sound playing as heard by the listener, and observed that the shading of the sound wave
developed higher contrast (corresponding to higher pressure fluctuations) and the sound
got louder.
(c) To produce a lower pitch tone, the speaker must oscillate back and forth more times
each second.
False. I determined this by observing that when I change the frequency slider, the rate
at which the speaker cone oscillates changes and the pitch I heard also changes. When I
decreased the frequency, the speaker oscillated fewer times per second and the pitch of
the tone was lower.
(d) When the speaker is producing sound waves, the motion of the speaker produces a slight
decrease in the pressure in front of the speaker during the forward motion of the speaker
head and a slight increase in pressure during the backward motion of the speaker head.
False. From the earlier part of this question, I determined that a lighter shade of gray
corresponds to an increase in pressure, and I observed that when the speaker moves
forward the shade of gray in front of it gets lighter. Therefore when the speaker moves
forward it produces a slight increase in pressure.
(e) To increase the volume of a tone at 400 Hz heard by the listener, the speaker must
oscillate back and forth fewer times each second than it does to produce the tone with
lower volume.
False. I determined this by playing with the frequency and amplitude controls. When
I change the frequency slider, the rate at which the speaker cone oscillates changes and
the pitch I heard also changes. When I decreased the frequency, the speaker oscillated
fewer times per second and the pitch of the tone was lower. Changing the rate at which
the speaker oscillated never changed the volume heard by the listener.
2. Suppose Prof. Betterton did a class demo and measured both frequency and wavelength of a
sound wave, and found them to be 180 Hz and 1.994 m, respectively. What speed of sound
does this imply? Discuss briefly at least two different reasons why this answer is not exactly
344 m/s, like the book says it should be (would the reasons you came up with make the
answer come out a little too high, low, or would it just be random? Why?)
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The speed of a wave with frequency f and wavelength λ is v = f λ. Here that implies v = 180
s × 1.994 m = 358.9 m/s.
The measured sound speed could be different because the temperature in the room could be
a bit warmer than standard air temperature—this would tend to make the measured speed
of sound a bit higher, but not enough to explain this measurement! If it were quite warm
in the lecture room (say 77 degrees F, which is 25 degrees C), the sound speed would only
increase to 347 m/s, still well below the measured value.
A more likely explanation for the difference between the measured value and the true value is
measurement error. I can’t measure the wavelength and the frequency with perfect precision.
If I had an error in either measurement of a few percent, that could explain why my measured
sound speed is off by about 4% from the true value.
3. In the following graph we have plotted the pressure measured at the listener’s ear as a function
of time for a 200 Hz tone generated by the speaker. In the answer graphs of pressure versus
time below, the dashed line indicates the original 200 Hz tone. (red dashed, on the computer
screen)
Look at each statement below, and decide which of the blue solid curves above represents the
variation in the pressure at the eardrum versus time.
(a) The volume of the speaker is turned up.
D. Increasing the volume makes the amplitude of the pressure wave higher without
changing the frequency.
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(b) The speaker is producing a lower pitch tone.
A. Playing a lower pitch decreases the frequency (increases the period), without changing
the amplitude.
(c) The speaker is oscillating back and forth, completing this cycle 400 times per second.
B. The original sound was at 200 Hz, so changing to 400 Hz doubles the frequency
(halves the period) without changing the amlitude. This means that one full cycle takes
0.0025 s, as shown in B.
(d) The speaker sound is unchanged, but the listener has moved farther from the speaker.
E. Moving farther from the speaker makes the sound heard softer, which corresponds
to a decrease in ampitude of the pressure wave at the listener’s ear without a change in
frequency.
4. As you sing a pure steady note of frequency 1700 Hz, there is a pressure wave traveling
outwards from you (at the speed of sound). What is the wavelength of that sound wave?
Assume the temperature is 20 o C, room temperature. Is this a high or low note? How does
that wavelength compare with ordinary sizes of things like doors, people’s heads, etc? What
is the longest and shortest wavelength that you could produce by singing? If the temperature
in the room was higher (but you sing the same pitch), does the wavelength change? If so,
how?
This calcualtion is based on v = f λ. Since we want to find the wavelength, rearrange to find
λ = v/f . Then
344 m/s
v
= 0.20 m.
λ= =
f
1700 /s
A wavelength of 0.2 m is about 8 inches—roughly the distance from my thumbtip to fingertip
if I spread my fingers out as wide as I can. This is a very high note: by comparing to the
chart in the back of the book I learned that 1700 Hz is close to G#6, towards the top of the
piano’s range and quite a bit higher than I could sing. To find out the highest/lowest pitch I
could sing, I used the PhET sound simulation, varied the frequency, and tried to match the
sound produced. 100 Hz was about the lowest I could do without sounding too awful, and
700 Hz was about the highest. If the temperature in the room goes up but I still sing the
same pitch, the wavelength has to go up, since v goes up and λ = v/f .
5. Long answer question: see comments above.
6. Go back to the simulation from question 1. Measure the speed of sound at two different
frequencies (200 Hz and 400 Hz). Briefly describe the process you used to find the speed and
then compare your answer to the speed of sound in air at sea level at 0 degrees C (331 m/s).
How different are the two speeds you come up with? Do you conclude that the speed of sound
depends on the frequency of the sound, or not? (Why?)
I did this by pausing the simulation, putting the ruler’s zero and the blue line right at the
edge between a light & dark band, starting the timer. Then I un-paused the simulation, let
it run until the specific edge between light & dark bands I was watching got to the 5 m mark
on the ruler and the other blue line, and paused it again. At 202 Hz, the simulation time was
0.0143 s for 5 m traveled. This gave me a speed estimate of v = d/t = 5 m/0.0143 s = 349.65
m/s.
I repeated the measurement at 404 Hz and found it took 0.0149 s for 5 m traveled, for a speed
estimate of 335.57 m/s.
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I think my measurement was pretty good, considering that I can’t read off the ruler or
pause/un-pause the simulation with perfect accuracy. My measurements at the two different
frequencies were different by about 4%, which is within my experimental error.
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