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# Simple Harmonic Motion Practice Problems for AP Physics B & C

based on 2 ratings
By — McGraw-Hill Professional
Updated on Feb 11, 2011

Review the following concepts if necessary:

1. A basketball player dribbles the ball so that it bounces regularly, twice per second. Is this ball in simple harmonic motion? Explain.

### Multiple Choice:

1. A pendulum has a period of five seconds on Earth. On Jupiter, where g – 30 m/s2, the period of this pendulum would be closest to
1. 1 s
2. 3 s
3. 5 s
4. 8 s
5. 15 s
2. A pendulum and a mass on a spring are designed to vibrate with the same period T. These devices are taken onto the Space Shuttle in orbit. What is the period of each on the Space Shuttle?
3. A mass on a spring has a frequency of 2.5 Hz and an amplitude of 0.05 m. In one complete period, what distance does the mass traverse? (This question asks for the actual distance, not the displacement.)
1. 0.05 cm
2. 0.01 cm
3. 20 cm
4. 10 cm
5. 5 cm
4. Increasing which of the following will increase the period of a simple pendulum?
1. the length of the string
2. the local gravitational field
3. the mass attached to the string
1. I only
2. II only
3. III only
4. I and II only
5. I, II, and III

### Free Response:

1. A mass m is attached to a horizontal spring of spring constant k. The spring oscillates in simple harmonic motion with amplitude A. Answer the following in terms of A.
1. At what displacement from equilibrium is the speed half of the maximum value?
2. At what displacement from equilibrium is the potential energy half of the maximum value?
3. When is the mass farther from its equilibrium position, when its speed is half maximum, or when its potential energy is half maximum?

### Solutions

1. The ball is not in simple harmonic motion. An object in SHM experiences a force that pushes toward the center of the motion, pushing harder the farther the object is from the center; and, an object in SHM oscillates smoothly with a sinusoidal position–time graph. The basketball experiences only the gravitational force, except for the brief time that it's in contact with the ground. Its position–time graph has sharp peaks when it hits the ground.
2. B—The period of a pendulum is
3. All that is changed by going to Jupiter is g, which is multiplied by 3. g is in the denominator and under a square root, so the period on Jupiter will be reduced by a factor of . So the original five-second period is cut by a bit less than half, to about three seconds.

4. A—The restoring force that causes a pendulum to vibrate is gravity. Because things float in the Space Shuttle rather than fall to the floor, the pendulum will not oscillate at all. However, the restoring force that causes a spring to vibrate is the spring force itself, which does not depend on gravity. The period of a mass on a spring also depends on mass, which is unchanged in the Space Shuttle, so the period of vibration is unchanged as well.
5. C—The amplitude of an object in SHM is the distance from equilibrium to the maximum displacement. In one full period, the mass traverses this distance four times: starting from max displacement, the mass goes down to the equilibrium position, down again to the max displacement on the opposite side, back to the equilibrium position, and back to where it started from. This is 4 amplitudes, or 0.20 m, or 20 cm.
6. A—The period of a pendulum is
7. Because L, the length of the string, is in the numerator, increasing L increases the period. Increasing g will actually decrease the period because g is in the denominator; increasing the mass on the pendulum has no effect because mass does not appear in the equation for period.

8.
1. The maximum speed of the mass is at the equilibrium position, where PE = 0, so all energy is kinetic. The maximum potential energy is at the maximum displacement A, because there the mass is at rest briefly and so has no KE. At the equilibrium position all of the PE has been converted to KE, so
2. 1/2kA2 = 1/2mv2max

Solving for vmax, it is found that

So, at a position 1/2 A, the velocity is half its maximum.

Now that we have a formula for the maximum speed, we can solve the problem. Call the spot where the speed is half-maximum position 2.

Write conservation of energy from the maximum displacement to position 2:

The speed v2 is half of the maximum speed we found earlier, or Plug that in and solve for x2:

The m's and the k's cancel. The result is or about 86% of the amplitude.

3. The total energy is 1/2kA2. At some position x, the potential energy will be 1/2 of its maximum value. At that point, 1/2kx2 = 1/2(1/2kA2). Canceling and solving for x, it is found that
4. This works out to about 70% of the maximum amplitude.

5. Since we solved in terms of A, we can just look at our answers to (a) and (b). The velocity is half-maximum at or 86%, of A; the potential energy is half-maximum at or 71%, of A. Therefore, the mass is farther from equilibrium when velocity is half-maximum.

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