Simple Harmonic Motion – MC Review KEY
1. A block attached to an ideal spring undergoes simple harmonic motion. The acceleration of the block has its maximum magnitude at the point where:
a. the speed is the maximum.
b. the potential energy is the minimum. c. the speed is the minimum.
d. the restoring force is the minimum. e. the kinetic energy is the maximum.
*C. The acceleration is a max. at the turnaround points---this is where the spring is maximally stretched or compressed and so the pull back to equilibrium is the strongest. By definition of a turnaround point, this is where the speed is zero.
2. A student measures the maximum speed of a block undergoing simple harmonic oscillations of amplitude A on the end of an ideal spring. If the block is replaced by one with twice the mass but the amplitude of its oscillations remains the same, then the maximum speed of the block will:
a. decrease by a factor of 4. b. decrease by a factor of 2. c. decrease by a factor of 2 d. remain the same.
e. increase by a factor of 2.
* C. Compare TME’s when all the energy is kinetic (at the equilibrium) and all of the energy is stored (at turnarounds): 2 of factor a by speed max the decrease would mass the doubling Thus, V : V for Solving 2 1 2 1 2 Max 2 2 max m KA KA MV
3. A linear spring of force constant K is used in a physics lab experiment. A block of mass m is attached to the spring and the resulting frequency, f, of the simple harmonic oscillations is measured. Blocks of various masses are used in different trials, and in each case, the corresponding frequency is measured and recorded. If f2 is plotted versus 1/m, the graph will be a straight line with slope of:
a. 2 2 4 K b. K 2 4 c. 42K d. 2 4 K e. 2 2 4 K
* D. Start with the formula for the period of a spring:
2 2 2 4 K : yields 1/m, out the Factor 4 f : graphed is match what to everything square Now, 2 1 f : everything flip so period, the of reciprical the is frequency The 2 m K m K K m T
4. A simple pendulum executes simple harmonic motion as it swings through small angles of oscillation. If max denotes the amplitude of the oscillations, which of the following statements is true?
a. When = 0, the tangential acceleration is zero. b. When = max, the tangential acceleration is zero. c. When = 0, the speed is zero.
d. When = 0, the restoring force is maximized. e. When = max, the speed is maximized.
*A. The tangential acceleration is caused by the component of weight that is tangent to the path (as opposed to the centripetal acceleration needed for the pendulum to travel in a circle). At the bottom of its path, the weight vector is acting radially down, which causes no acceleration. At any other position, a component of the weight (mgsin ) is acting tangent to the path, causing the pendulum to speed up or slow down.
5. When a mass
m
is hung on a certain ideal spring, the spring stretches a distanced
. If the mass is then set oscillating on the spring, the period of oscillation is proportional to:a. g d b. d g c. mg d d. d g m2 e. g m
* A. This problem combines two concepts: the formula for period and the definition of equilibrium for a vertical spring: g d K m T 2 d mg m 2 T : this ng Substituti d mg K so mg Kd spring the of force upward the Thus, : d distance a stretches it spring on the hangs mg of weight a when fact that the use : K find To 2
6. A particle moves in a circle in such a way that the X- and Y-coordinates of its motion are given in meters as functions of time t in seconds by:
X = 5 cos (3t) Y = 5 sin (3t)
What is the period of revolution of the particle? a. 1/3 second. b. 3 seconds. d. seconds 3 2 e. seconds 2 3 f. 6 seconds. * D.
First off, don’t be confused by the fact that there is an X and a Y function.
Remember the phonograph demo. Whether you just looked at the horizontal or just looked at the vertical position, either one would oscillate in simple harmonic motion, so just focus on one.
Remember that the formula for SHO is X = A cos ( t)
So, the rotational speed is 3 radians/sec. We need to “convert” this to the time to complete one revolution:
3 2 revolution radians 2 3radians 1second T
7. The displacement vs. time for a particle in simple harmonic motion is shown above. Which of the following graphs shows the kinetic energy, K, of the particle as a function of time, t, for one cycle of motion?
a.
b.
c.
d.
* B
First off, you know the KE can never be negative, which throws out graphs A, D, and E.
Second, KE is proportional to velocity squared. So, think about taking the derivative of the position graph and then squaring it. This means max’s on the position graph become zeros of the KE graph. This should square with what you know about SHO—max positions are turnaround points = no KE.
8. The length of a simple pendulum with a period on Earth of one second is most nearly: a. 0.12 meters. b. 0.25 meters. c. 0.50 meters. d. 1.0 meters. e. 10.0 meters. *B. .247 is answer the anything, e approximat t don' you if : Note 25 . 40 10 36 10 L sides both square 10 L 36 1 3 about is 3.14 say 10m/s L 2(3) sec 1 2 2 g L T
9. A ball is dropped from a height of 10 meters onto a hard surface so that the collision at the surface may be assumed elastic. Under such conditions the motion of the ball is:
a. simple harmonic with a period of about 1.4 seconds. b. simple harmonic with a period of about 2.8 seconds. c. simple harmonic with an amplitude of 5 m.
d. periodic with a period of about 2.8 seconds but not simple harmonic. e. motion with constant momentum.
* D. The motion is not simple harmonic because the position function of free fall is parabolic (constant acceleration), not sinusoidal (acceleration that decreases as object heads towards equilibrium). If you use kinematics to find time to fall you get 1.41 seconds, so the time to return to original position is about 2.82 seconds. The collision must be elastic or the object will not return to its original position.
Questions 10 and 11:
10. (1984) A 0.1 kg block is attached to an initially unstretched spring of force constant K= 40 N/m as shown above. The block is released from rest at time t = 0. What is the amplitude of the resulting simple harmonic motion of the block?
a. m 40 1 b. m 20 1 c. m 4 1 d. m 2 1 e. 1m
* A. The release point is the top turnaround point. The difference between this point and equilibrium is the amplitude: m A s m kg A m N mg x K m Equilibriu 40 1 ) / 10 )( 1 (. ) )( / 40 ( : 2
11. (1984) At what time after release will the block first return to its initial position? a. s 40 b. s 20 c. s 10 d. s 5 e. s 4 *C
This is the definition of one full period:
10 20 2 400 1 2 / 40 1 . 2 2 m N kg K m T
12. (1993) A simple pendulum consists of a 1 kg brass bob on a string about 1.0 meters long. It has a period of 2.0 seconds. The pendulum would have a period of 1.0 second if the:
a. string were replaced by one about 0.25 meters long. b. string were replaced by one about 2.0 meters long. c. bob were replaced by a .25 kg brass sphere. d. bob were replaced by a 4.0 kg brass sphere. e. amplitude of the motion were increased. * A
They give you a lot of excess information here to trick you. Remember that the period of a pendulum depends only on its length and the planet it is on. You can check that a pendulum on Earth of length 1 meter does have a period of about 2 seconds. To cut its period in HALF, you would need to reduce the length by a factor of FOUR, since it is under a square root in the period formula.
13. (1974) The forces produced by two springs as they are stretched are shown in the graph above. Spring 1 indicated by the dashed line is linear (Hookian), but spring 2 is not. The period of oscillation for a mass attached to spring 2 is:
a. dependent on amplitude but never greater than for spring 1. b. dependent on amplitude but never less than for spring 1. c. dependent on amplitude but always equal to that of spring 1. d. independent of amplitude and never greater than for spring 1. e. independent of amplitude and never less than for spring 1. *A.
This one is tricky. Because the spring is nonlinear, the solution to the differential equation is not a sine function and the period does depend on the amplitude. If you look at the graph closely, you will see that the FORCE of spring 2 increases with displacement faster than the force for spring 1. It should make sense that if the restoring force is GREATER than it would be for standard simple harmonic motion, the spring will pull the mass back to equilibrium more quickly, reducing the period.
14. (1974) An object is suspended from a spring whose mass is negligible compared to that of the object. The object is displaced slightly, and the period of its motion is observed to be T seconds. The spring is then cut in half and the object is suspended from one of the halves. The object is displaced slightly and its period is observed to be T’ seconds. The ratio T’/T is:
a. ½ b. 2 1 c. 1 d. 2 e. 2 * B
2 1 1 K m 2 2 2 T T : is ratio the So, 2K m 2 T ht. given weig a for stretch total less so coils, fewer are there becuase constant spring the DOUBLES actually half in spring the Cutting 2 2 1 K m K m T15.
(2004, 46 %)
A simple pendulum has a period of 2 seconds for small amplitude oscillations. The length of the pendulum is most nearly:
a. 1/6 meters. b. ¼ meters. c. ½ meters. d. 1 meters. e. 2 meters. * D 1 10 L 10 1 10 L 1 10 L 2 2 g L 2 2 2 L T
