Physics 1240 Website: http://www.colorado.edu/physics/phys1240
Homework 1 Physics 1240 Spring 2005
Due: Thursday 20 January
In all homework problems where there are calculations, you should show the steps you
took in arriving at your result: the correct answer alone is not enough! Also be sure to
include units with any results, e.g. “v = 330 m/s” (not just “v = 330”).
1. How many meters does sound travel in one minute? How far in an hour? How far
in 1 millisecond (0.001s)? Assume the air is at 20ºC.
Sound travels 344 m in 1 sec. There are 60 secs in a minute and 3600
secs in an hour. The distance traveled is the product d= vt:
in 1 min: d = 344x60 = 20,640 m (more than 20 km)
in 1 hour: d = 344x3600 = 1238.4 km (about 774 miles)
in 1 ms: d = 344x0.001 = 0.34 m (which is about 1 ft)
2. How long does the sound of a starter’s pistol take to travel to the finish line of the
100m dash on a hot day when the temperature is 30ºC?
Sound travels at 344 m/s at 20ºC but the speed increases when it’s
hotter. Since the increase in v is +0.6 m/s per ºC, the increase
corresponding to a 10ºC temperature difference is 6 m/s, meaning
that at 30ºC sound travels at 350 m/s.
3. The human ear can detect sounds in the approximate frequency range 20 Hz to 20
kHz. What are the largest and smallest wavelengths the ear can detect?
Inverting the relation v = λf gives
at 20 Hz : λ = 17.2 m
at 20 kHz : λ = 17.2 mm
4. If the frequency of a sound wave is 440Hz (corresponding to the A above middle
C) what is the period of each vibration?
since the period T is the inverse of the frequency f,
T = 1/f = 2.27 ms
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Physics 1240 Website: http://www.colorado.edu/physics/phys1240
5. Look at Figs. 2.8 and 2.9 in Hall. Identify another situation from everyday life
that illustrates a functional relationship and sketch a graph that represents it.
This graph shows the variation of air pressure with time over the last
week measured at NCAR. Note that 1 pascal = 1 Pa = 1 N/m2.
6. Ultrasound with a frequency of 4.25 MHz can be used to image internal organs of
the human body. If the speed of sound in the body is the same as in salt water (1.5
km/s), what is its wavelength in the body? Does this seem like a good wavelength
to resolve small objects, such as a growing fetus?
v = fλ, or λ = v/f
f = 4.25 x 106 Hz, or 4,250,000 Hz
v = 1500 m/s
λ = v/f = 1500/4,250,000 = 0.00035 m
This is an extremely small wavelength. Small enough that it passes
through most fluids, but reflects off most tissues, yielding a “sonar”
picture of a fetus.
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Physics 1240 Website: http://www.colorado.edu/physics/phys1240
7. Of the four microphone positions A, B, C, and D shown in the Figure, which will
best pick up the sound of the tuning fork (i.e. where is the sound loudest)? Why?
The tuning fork is usually
struck on the side, from
the direction of
microphone A. The tuning
fork has its largest natural
response in this direction.
The biggest vibrations of the tines are consequently also in the
direction of microphone A so this is where the loudest sound will be
detected. However, the whole tuning fork does vibrate and so some
sound can be picked up by any of the microphones.
8. If the rarefactions of a particular sound wave reduce the pressure to 0.997 atm,
what is the pressure in the compressions? What is the amplitude of this wave,
expressed first in atm and then in N/m2? Is this a soft sound or a loud sound?
Assuming that ambient (background) pressure is 1 atm = 105 N/m2, the
amplitude of the fluctuations is 0.003 atm or 300 N/m2. The pressure
in the compressions is (1 + 0.003) atm = 1.003 atm.
This pressure wave corresponds to a very loud sound (comfortable
hearing levels corresponding to much smaller pressure fluctuations, in
the range 10-7 - 10-5 atm).
9. You see lightning strike, then hear thunder 5 s later. How far away was the strike?
v = d/t => d = vt
v = 344 m/s
t=5s
d = vt = (344 m/s)(5 s) = 1720 m
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Physics 1240 Website: http://www.colorado.edu/physics/phys1240
10. Doppler Shift: Describe what you hear if you are in a train traveling rapidly past a
stationary bell ringing at a railroad crossing and why.
As you approach the bell on the train, the pitch will remain relatively
constant. When you pass the bell, the pitch of the bell drops, then
remains at that lower pitch as long as you are moving at a constant
velocity.
As you move towards the bell, you are speeding up the number of
compressions and rarefactions that go past you per second. In other
words, the waves are going past you more frequently, thus the
“frequency” increases, and the apparent pitch is higher.
When you pass the bell, exactly the opposite occurs. The waves are
moving in the same direction as you, so their apparent speed is slower,
and the waves move past you less frequently. Less frequency means lower
pitch.
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