Physics 53
Summer 2008
Exam III
Solutions
In questions or problems not requiring numerical answers, express the answers in terms of the
symbols given, and standard constants such as g. If numbers are required, use g = 10 m/s2.
Part A: Multiple choice questions. Check the best answer. Each question has a value of 4 points.
1.
A system’s motion is described by the equation x(t) = Acos ! t . Which of the
following is NOT true?
If A is independent of ! the motion is simple harmonic motion.
The system starts at t = 0 with zero velocity.
The potential energy can be written as U(x) = 12 m! 2 x 2 .
√
2.
The total force on the system is F = m! 2 x . [Minus this.]
For the average input power Pav for a driven oscillator with damping, which of
the following is NOT true?
The maximum value of Pav occurs when the frequency of the driving
force is equal to the frequency at which the oscillator would oscillate
freely without damping.
√
If the damping is small, the maximum value of Pav is less than if the
damping is large. [Larger.]
The curve of Pav vs. ω is more sharply peaked if the damping is small
than if it is large.
Pav multiplied by the period of the oscillation is equal to the energy
dissipated by the damping in one complete oscillation.
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3.
Summer 2008
Two harmonic waves of the same wavelength and frequency are interfering.
Their individual intensities are I0 and 4 I0 .
The minimum that the resulting intensity can be is 3 I0 .
The second wave has 4 times the amplitude of the first wave.
√
The maximum that the resulting intensity can be is 9 I0 . [The second wave
has twice the amplitude of the first. Constructive interference makes the
amplitude three times that of the first wave, so the intensity is 9 times.]
None of the above is true.
4.
On the piano strings for different low notes have equal length and tension, while
strings for different high notes have equal mass per unit length and tension.
Which of the following is NOT true?
√
The strings for two low notes producing 55 Hz and 110 Hz have mass per
unit lengths in the ratio 2:1. [ 2 : 1 .]
Strings for high notes producing 440 Hz and 880 Hz have lengths in the
ratio 2:1.
If the string for 440 Hz is 1 m long, then if the string for 55 Hz were made
of the same wire and had the same tension it would be 8 m long.
One of the three strings for 440 Hz is out of tune, producing 3 beats/s
when the note is played; the bad string’s frequency might be 437 Hz.
Part B: True-false questions. Check the correct answer. Each question has a value of 3 points.
1.
The period of the swing of a simple pendulum is independent of the angle
through which it swings.
True
√
2.
False [Only approximately, for small swings.]
Water flowing from a hose through the constriction of a nozzle has lower
pressure than the water in the hose.
√
True [Continuity.]
False
3.
If sound waves from two sources interfere destructively in some region of space
they will interfere constructively in some other region.
√
True [Energy is conserved]
False
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Physics 53
Summer 2008
Part C: Problems. Work problems in the space provided, indicating your method clearly. A correct final
answer supported by no argument, or a fallacious one, will receive little or no credit. Problems carry the
point values shown.
1.
Two questions about fluids, 10 points each:
a.
A cylindrical block of wood of cross-section area A floats upright in water,
with its bottom at depth h below the surface. Show that if it is in an
environment where g is different (but not zero) it will float at the same
level. That is, prove that the value of h is independent of g.
The volume of water displaced is Ah so the buoyant force is Ah! g , where
! is the density of water. This equals the weight because the block floats,
so Ah! g = mg , where m is the mass of the block. We find h = m/ A! ,
which is independent of g as claimed.
b.
A large tank has a hole in its side as shown with
a pipe which directs the emerging stream
directly upward. Prove that the water in the
emerging stream rises to the same height as the
level of liquid in the tank.
Let the top of the water in the tank be at height h above the opening of the
pipe. Use Bernoulli to compare those two points: P0 + ! gh = P0 + 12 ! v 2 , so
v 2 = 2gh . Now use conservation of energy to find the height h’ at which
the stream is momentarily at rest:
h! = h as claimed.
3
1
mv 2
2
= mgh! . This gives v 2 = 2gh! , so
Physics 53
2.
Summer 2008
Two identical small blocks of mass m are as shown, with one on top of the other
and the lower block attached to a spring of stiffness k. The coefficient of static
friction between the blocks is µs . They are oscillating with amplitude A.
a.
Assuming the top block does not slip on the bottom one, what is the
angular frequency ω of the oscillation?
b.
What is the largest value of A for which the top block will not slip at any
point in the motion? Answer in terms of given quantities.
c.
Suppose A is slightly larger than this value so the top block slides off
when the spring is maximally stretched. Neglect energy lost to friction
and find the maximum speed of the bottom block in its new oscillation.
Answer in terms of given quantities.
[15 points]
a.
The mass is 2m so we have ! = k / 2m .
b.
The maximum acceleration friction can provide for the top block occurs
when the friction force has its maximum value, so mamax = µs N = µs mg , or
amax = µs g . The largest acceleration during the oscillation is ! 2 A , so the
largest value of A for which the top block will not slip is given by
! 2 A = µs g , or A = µs g /! 2 = µs g " (2m/ k) .
c.
If the block slips off at that point both blocks are momentarily at rest, so
there is no loss of energy to the oscillation when this happens. The
amplitude remains A, but the angular frequency becomes " ! = k /m . The
maximum speed in the oscillation is vmax = " !A = A k /m . Since the
amplitude is essentially the same as in (b) we find vmax = 2 µs g m/ k .
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Physics 53
3.
Summer 2008
The “Cartesian diver” is a small glass cone open at the
bottom which has a pocket of air trapped inside. It is in
a sealed flexible plastic container nearly filled with
water, and initially the diver is floating at the surface as
shown. When one squeezes the sides of the container,
reducing the volume of the air above the water, the
diver sinks toward the bottom. When the squeezing
force is removed the diver rises again and floats.
Explain this behavior in some detail. [What happens to
the average density of the diver, and why?]
Diver
[10 points]
When the air above the water is compressed, the rise in air pressure results in
increased pressure at all points in the water (Pascal’s law), so the water inside the
diver is pushed up to compress the air in it. This allows more water to flow into
the diver, increasing its average density. When this density exceeds that of water,
the diver sinks. Removing the extra pressure on the container allows the extra
water to flow out of the diver, reducing its average density to its original value,
so it floats again. One can make it hover at any depth by making its average
density equal to that of water.
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Physics 53
4.
Summer 2008
In a radar speed detection system a stationary police car sends out radar waves
of frequency f0 = 109 Hz traveling at speed c = 3 ! 108 m/s. This wave reflects
from a moving car and the reflected wave is detected by the police car’s system.
The beats between outgoing and incoming waves determine the speed of the car.
The car is approaching the police car at speed v = 30 m/s. Let x = v /c , which is a
small number, and use the ordinary formulas for the Doppler effect.
a.
What is the ratio f1 / f0 , where f1 is the radar frequency reflected by the
car? (Give the answer in terms of x.)
b.
What is the ratio f 2 / f0 , where f 2 is the frequency received back by the
police car? (Give the answer in terms of x.)
c.
What is the beat frequency? (Give the answer numerically.)
[Use the binomial approximation (1 ! x)!1 " 1 + x where appropriate.]
[15 points]
a.
Here the source (police car) is at rest and the receiver (car) is moving
f
c+v
toward it. We have f1 = f0
= f0 (1 + x) , so 1 = 1 + x .
c
f0
b.
Now the source (car) moves toward the receiver (police car) at rest. We
f
c
1
1+ x
1+ x
have f 2 = f1
, so 2 =
.
= f1
= f0
c!v
1! x
1! x
f0 1 ! x
c.
2x
"1+ x %
The beat frequency is f 2 ! f0 = f0 $
! 1' = f0
( f0 ) 2x , since
#1! x &
1! x
x = 10!7 . We find for the beat frequency 109 ! 2 " 10#7 = 200 Hz.
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Physics 53
5.
Summer 2008
Two questions about standing waves:
a.
Bugle calls use frequencies in the ratios 1 : 4/3: 5/3 : 2. The lowest of these
is 300 Hz and they are sequential harmonics. What number harmonics are
they, and what is the bugle’s fundamental frequency?
[10 points]
The frequencies are 300, 400, 500 and 600 Hz. The difference between
successive frequencies is 100 Hz. Since each frequency is divisible by this,
it is the fundamental frequency. The harmonic numbers are 3, 4, 5 and 6.
b.
Write wavefunctions for two waves which, when superposed, will give a
standing wave. The waves should each have amplitude 0.01 m and speed
10 m/s. The nodes in the standing wave should be 0.1 m apart.
[5 points]
We have y1 = 0.01cos(kx ! " t), y2 = 0.01cos(kx + " t) , where
k = 2! / " = 2! /0.2 = 10! (because the nodes are separated by ! / 2 ).
Also. The speed is ! / k , so ! = 10k = 100" . The final results are
y1 = 0.01cos(10! x " 100! t), y2 = 0.01cos(10! x + 100! t) .
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Physics 53
Summer 2008
Median = 71.5
SD = 13.7
Median = 60.0
SD = 13.7
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