Page 187 - Schaum's Outline of Theory and Problems of Applied Physics

172 SIMPLE HARMONIC MOTION [CHAP. 14

SOLVED PROBLEM 14.12
An object suspended from a spring oscillates at 10 Hz with an amplitude of 5 cm. (a) What is the total
energy of the motion? (b) What change in the amplitude would double the energy?
(a) The total energy of the motion is, since A = 0.05 m,
2
2
2
2
2
2
E = 2π f A = 2π (10 Hz) (0.05 m) = 4.9J
2
2
2
(b) Since E is proportional to A , to double E means increasing the amplitude from A 1 to A 2 so that A = 2A .
1
2
Hence the new amplitude should be
√ √

2
A 2 = 2A = 2A 1 = 2(5cm) = 7cm
1
PENDULUMS
A simple pendulum has its entire mass concentrated at the end of a string, as in Fig. 14-3(a), and it undergoes
simple harmonic motion provided that the arc through which it travels is only a few degrees. The period of a
simple pendulum of length L is

L
T = 2π simple pendulum
g
Pivot
O
h

L

Center of gravity q
m

(a) (b) (c)
Fig. 14-3

The physical pendulum of Fig. 14-3(b) is an object of any kind which is pivoted so that it can oscillate freely.
If the moment of inertia of the object about the pivot O is I, its mass is m, and the distance from its center of
gravity to the pivot is h, then its period is

I
T = 2π physical pendulum
mgh
A torsion pendulum consists of an object suspended by a wire or thin rod, as in Fig. 14-3(c), which undergoes
rotational simple harmonic oscillations. From Hooke’s law, the torque τ needed to twist the object through an
angle θ is

τ = Kθ
provided the elastic limit is not exceeded, where K is a constant that depends on the material and dimensions of
the wire. If I is the moment of inertia of the object about its point of suspension, the period of the oscillations is

I
T = 2π torsion pendulum
K