Download Physics 262-005 23 October, 2000 EXAMINATION II SOLUTIONS

Physics 262-005
23 October, 2000
EXAMINATION II SOLUTIONS (WHITE)
1. A wave is described by y(x,t) = 0.1 sin(3x - 10t), where x is in meters, y is in centimeters
and t is in seconds. The frequency is:
A) 20 Hz
B) 10 Hz
C) 10= Hz
D) 5= Hz
E) 20= Hz
In this problem, 0:1 is the amplitude, k = 3 (the angular wave number) and ! = 10 rad/s
(the angular frequency). Then
f = 2! = 5
The correct answer is D
2. Below are sets of values for the spring constant, damping constant, and mass for a particle
in damped harmonic motion. Which of the sets takes the longest time for its mechanical
energy to decrease to one-fourth of its initial value?
A) k0; b0; 10mo
B) k0=2; 6b0; 2mo
C) 3k0 ; 2b0; mo
D) 4ko ; b0; 2mo
E) k0; b0; mo
The damping time is set by , = b=m and, as shown in the formula page, by exp , (,t=2).Large
values of , mean rapid damping. Therefore, we are looking for the SMALLEST value of ,
among the various choices. In A), , = b=10m, in B) , = 3b=m, in C) , = 2b=m, in D)
, = b=2m, and in E) , = b=m. Clearly in case B) , has the largest value and in case A)
, has the smallest value! The larger the value of ,, the more rapidly the energy dissipates.
Conversely, the smaller the value of ,, the more slowly the energy dissipates. Therefore, the
correct answer is A.
3. Two events occur 100 m apart with an intervening time interval of 0.60 s. The speed of
a reference frame in which they occur at the same coordinate is:
A) 0
B) 0.25c
C) 0.56c
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D) 1.1c
E) 1.8c
Here we apply the Lorentz transformation for x0 such that
x0 = (x , ct)
For x0 = 0 we have x = ct or 100 = 6 10,7 c = 180 . Then it is easy to see that = :56
The correct answer is C.
4. A wave traveling to the right on a stretched string is shown below. The direction of the
instantaneous velocity of the point P on the string is:
A) "
B) #
C) ,!
D) %
E) no direction since v = 0
This is quite hard. The wave moves to the right. The easiest thing to do is draw the wave at
a later time, in which case we see that the point P has moved DOWN. Therefore, the point
P is going to have a smaller value at later times than it has at the time shown. This means
that the point P is moving DOWN. The correct answer is B.
5. A rocket ship of rest length 100 m is moving at speed 0.8c past a timing device which
records the time interval between the passage of the front and back ends of the ship. This
time interval is:
A) 0.20 s
B) 0.25 s
C) 0.33 s
D) 0.52 s
2
E) 0.69 s
There are several ways to do this problem. The easiest, perhaps, is to realize that, in our
frame, the length of the rocket ship is 100= . So we have to nd when = 0:8 But this
is not hard, and we get (plugging in the numbers) that = 1:67 Thus the 100 meter rocket
ship looks, to us, as if it is L = 60 meters long. But 60 meters at speed v = 0:8c gives us a
time of :25 10,6 seconds. The correct answer i s B.
6. Three traveling sinusoidal waves are on identical strings, with the same tension. The
mathematical forms of the waves are y1(x; t) = ymsin(3x , 6t), y2(x; t) = ymsin(4x , 8t), and
y3(x; t) = ymsin(6x , 12t), where x is in meters and t is in seconds. Match each mathematical
form to the appropriate graph below.
A)y1 = i; y2 = iii; y3 = ii
B) y1 = i; y2 = ii; y3 = iii
C) y1 = iii; y2 = ii; y3 = i
D) y1 = iii; y2 = i; y3 = ii
E) y1 = ii; y2 = i; y3 = iii
In these diagrams, for which you needed a magnifying glass to see carefully, we are looking
at a snapshot in time. Time is frozen so we are looking at the wavelengths. We can see the
wavelengths from the three equations. In i, k = 3 so = 2:1 meters. Similarly, for the next
two diagrams, = 1; 57 and 1:05 meters respectively. The shortest wavelength is y3 and the
longest wavelength is y1. This means y2 is in the middle. When we look at the diagrams,
we see the longest wavelength is i and the shortest is iii. So we identify y1 with i and y3
with iii. The correct answer is B.
7. In the diagram below, the interval PQ represents:
A) wavelength/2
B) wavelength
C) 2 x amplitude
D) period/2
E) period
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This is meant to be a simple problem in identication. The interval shown is the period, the
distance between identical time points on the waveform. The correct answer is E.
8. A sinusoidal wave is traveling toward the right as shown. Which letter correctly labels
the wavelength of the wave?
A) A
B) B
C) C
D) D
E) E
Here we have a term identication problem again. The wavelength is A. so the correct answer
here is A.
9. A particle is in simple harmonic motion along the x axis. The amplitude of the motion is
xm. When it is at x = 1=2xm , its kinetic energy is K = 5J and its potential energy (measured
with U = 0 at x = 0) is U = 3J. When it is at x = xm, the kinetic and potential energies
are:
A) K = 5J and U = -3J
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B) K = 0 and U = 8J
C) K = 8J and U = 0
D) K = 0 and U = -8J
E) K = 5J and U = 3J
The total energy of our oscillator is 8 Joules. It will be a constant. Therefore, when x = xm
and the object is instantaneously stopped, we have KE = 0 and therefore PE = 8 Joules.
The correct answer is B.
10. In a photoelectric eect experiment the stopping potential is:
A) the energy required to remove an electron from the sample
B) the kinetic energy of the most energetic electron ejected
C) the potential energy of the most energetic electron ejected
D) the photon energy
E) the electric potential that causes the electron current to vanish
This is a matter of denition. Stopping potential is just that, the potential at which the
electron current STOPS. The correct answer is E.
11. A mass on a spring is subjected to a damping force that is proportional to its velocity
and to an applied sinusoidal force. The energy dissipated by damping is supplied by:
A) friction
B) the kinetic of the mass
C) the applied force
D) gravity
E) the potential energy of the spring
Energy can only come from an external source. The key words are \supplied by". The only
external source is the applied force. The correct answer is C.
12. A sinusoidal force with a given amplitude is applied to an oscillator. At resonance the
amplitude of the oscillation is limited by:
A) the damping force
B) the initial velocity
C) the force of gravity
D) the initial amplitude
E) none of the above
We have seen that the amplitude at resonance without damping is innite. Therefore, it
is damping (always present in reality) that prevents the amplitude from being innite. The
correct answer is A.
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13. A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m.
If the system has an energy of 6.0 J, then the amplitude of the oscillation is:
A) 0.06 m
B) 0.17 m
C) 0.24 m
D) 4.9 m
E) 6.9 m
At the end of its motion, when x = A all of the energy is potential. Then
E = PE = 12 kA2
Therefore we can solve and get A = 0:24 (we must not forget that the arithmetic involves
taking a square root of A2). The correct answer is C.
14. A meson when at rest decays 2 s after it is created. If moving in the laboratory at
0.99c, its lifetime according to laboratory clocks would be:
A) 4.6 s
B) 14 s
C) the same
D) 0.28 s
E) none of these
We always need to get from . This is not hard and the arithmetic leads us to = 7:09.
But this is the factor by which the lifetime is dilated so that, for us, t = 7:09 2 = 14 s.
The correct answer is B.
15. A photon in light beam A has twice the energy of one in light beam B. The ratio A/B
of the wavelengths is:
A) 1/2
B) 1/4
C) 1
D) 2
E) 4
We know that
E = hf = hc
Therefore, the ratio of wavelengths will be inverse to the ratio of energies. Since the ratio of
energies is TWO, the ratio of wavelengths is ONE HALF :The correct answer is A.
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16. An observer notices that a moving clock runs slow by a factor of exactly 10. The speed
of the clock is:
A) 0.100c
B) 0.0100c
C) 0.990c
D) 0.900c
E) 0.995 c
This is easy. The dilation factor here is = 10. We can solve for easily from this (careful
with the arithmetic though) and get 2 = 0:99 and = 0:995The correct answer is E.
17. A measurement of the length of an object that is moving relative to the laboratory
consists of noting the coordinates of the front and back:
A) at the same time according to clocks at rest in the laboratory
B) at the same time according to clocks that move with the object
C) at dierent times according to clocks at rest in the laboratory
D) at the same time according to clocks at rest with respect to the xed stars
E) none of the above
We must measure the front and back of the stick at the same moment, the same instant of
time, in our reference frame - i.e. the laboratory. Therefore, the correct answer is A.
18. In the accompanying gure all strings are the same except for string D which has a
greater linear density. Which of the answers below is true? (The tensions are TA; TB; TC and
TD )
A) TA = TB = TC = TD
B) TA = TB and TC = TD = 2TA
C) TB = 2TA and TC = TD = TA=2
D) TA = TB and TC = TD = TA=2
E) None of the above is true.
This is a basic mechanics problem. The tensions in A and B are the same because the wall
supplies the same force as the left-hand weight in B . The tensions in C and D are also
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the same but they are 1/2 of the values in A and B . This is due to the fact that the same
weight is supported by TWO strings and not by one string. Therefore, the correct answer is
D.
19. Consider two strings, A and B with exactly the same linear densities. String A has
twice the tension of string B . The ratio of the velocities of a wave propagated on string A
to that of string B (i.e. vA=vB ) is
A) 2
B) 1/2
p
C) 2
p
D) 1= 2
E) 4
This problem is related to the problem with weights and string densities above. If the densities
are the same and we know that the velocity is given by
s
v = T
Therefore, the ratio of velocities is the SQUARE
p ROOT of the ratio of the tensions. Since
TA = 2TB we have the ratio of the velocities is 2. The correct answer is C.
20. Consider two strings, A and B with exactly the same tensions The linear density of
string A is twice that for string B:The ratio of the velocities of a wave propagated on string
A to that of string B (i.e. vA=vB ) is
A) 2
B) 1/2
p
C) 2
p
D) 1= 2
E) 4
Here we have the linear densities that are dierent. But, for velocity, linear density occurs
in the denominator. Therefore the higher the linear density, thepLOWER the velocity. The
ratio of linear densities is TWO so the ratio of velocities is 1= 2. The correct answer is
D.
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