Page 189 - Schaum's Outline of Theory and Problems of Applied Physics

174 SIMPLE HARMONIC MOTION [CHAP. 14

From Fig. 10-3 the moment of inertia of the disk is

2
1
1
2
I = MR = ( )(2kg)(0.15 m) = 0.0225 kg·m 2
2 2
The period of oscillation is therefore

I 0.0225 kg·m 2
T = 2π = 2π = 0.482 s
K 3.82 N·m/rad
The frequency is f = 1/T = 1/0.482 s = 2.07 Hz.
SOLVED PROBLEM 14.18

To determine its moment of inertia about a diameter, a brass hoop is suspended by a wire whose torsion
constant is K = 25 N·m/rad. The hoop executes 6 oscillations per second. What is the moment of inertia?
√
2
2
From the formula T = 2π I/K for the period of a torsion pendulum we obtain I = T K/(4π ). Since
T = 1/f , the moment of inertia of the hoop is
K 25 N·m/rad 2
I = = = 0.0176 kg·m
2
−1 2
2
4π f 2 (4π )(6s )
Multiple-Choice Questions

14.1. The period of a simple harmonic oscillator does not depend on which one or more of the following?
(a) its mass (c) its force constant
(b) its frequency (d) its amplitude
14.2. The period of a simple pendulum depends upon its
(a) mass (c) maximum velocity
(b) length (d) total energy

14.3. When an object that is undergoing simple harmonic motion passes through its equilibrium position, its velocity is
(a) zero (c) its maximum value
(b) half its maximum value (d) none of the above

14.4. A spring is cut in three equal parts. If its original force constant was k, each new spring has a force constant of
(a) k/3 (c) 3k
(b) k (d) 9k

14.5. A vertical spring 60 mm long resting on a table is compressed by 5.0 mm when a 200-g mass is placed on it. The
force constant of the spring is

(a) 1.6 N/m (c) 196 N/m
(b) 40 N/m (d) 392 N/m

14.6. A spring is stretched by 30 mm when a force of 0.40 N is applied to it. The potential energy of the stretched spring is
(a) 2.8 × 10 −5 J (c) 6.0 × 10 −3 J
(b) 7.2 × 10 −5 J (d) 1.2 × 10 −2 J