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) Hz

B) Hz

C) Hz

D) Hz

E) Hz

In this problem, is the amplitude, k=3\ (the angular wave number) and rad/s (the angular frequency). Then

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)

B)

C)

D)

E)

The damping time is set by and, as shown in the formula page, by .Large values of mean rapid damping. Therefore, we are looking for the SMALLEST value of among the various choices. In A), , in B) , in C) , in D) , and in E) . 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

D) 1.1c

E) 1.8c

Here we apply the Lorentz transformation for \ such that

For we have or . Then it is easy to see that 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

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 . So we have to find when But this is not hard, and we get (plugging in the numbers) that Thus the 100 meter rocket ship looks, to us, as if it is L=60 meters long. But 60 meters at speed gives us a time of seconds. The correct answer is B.

6. Three traveling sinusoidal waves are on identical strings, with the same tension. The mathematical forms of the waves are , , and , where x is in meters and t is in seconds. Match each mathematical form to the appropriate graph below.

A)

B)

C)

D)

E)

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 meters. Similarly, for the next two diagrams, and meters respectively. The shortest wavelength is \ and the longest wavelength is . This means \ 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 \ with i and 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

This is meant to be a simple problem in identification. 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 identification 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 . When it is at , its kinetic energy is K = 5J and its potential energy (measured with U = 0 at x = 0) is U = 3J. When it is at , the kinetic and potential energies are:

A) K = 5J and U = -3J

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 and the object is instantaneously stopped, we have KE=0 and therefore PE=8 Joules. The correct answer is B.

10. In a photoelectric effect 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 definition. 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 infinite. Therefore, it is damping (always present in reality) that prevents the amplitude from being infinite. The correct answer is A.

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

Therefore we can solve and get (we must not forget that the arithmetic involves taking a square root of ). 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 . But this is the factor by which the lifetime is dilated so that, for us, 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

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 HALFThe correct answer is A.

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 . We can solve for easily from this (careful with the arithmetic though) and get and The 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 different times according to clocks at rest in the laboratory

D) at the same time according to clocks at rest with respect to the fixed 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 figure 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 and )

A)

B) and

C) and

D) and

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 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 to that of string B (i.e. ) is

A) 2

B) 1/2

C)

D)

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

Therefore, the ratio of velocities is the SQUARE ROOT of the ratio of the tensions. Since we have the ratio of the velocities is . 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 The ratio of the velocities of a wave propagated on string to that of string B (i.e. ) is

A) 2

B) 1/2

C)

D)

E) 4

Here we have the linear densities that are different. But, for velocity, linear density occurs in the denominator. Therefore the higher the linear density, the LOWER the velocity. The ratio of linear densities is TWO so the ratio of velocities is . The correct answer is D.

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