Page 181 - Schaum's Outline of Theory and Problems of Applied Physics

CHAPTER 14

Simple Harmonic

Motion

RESTORING FORCE

When an elastic object such as a spring is stretched or compressed, a restoring force appears that tries to return
the object to its normal length. It is this restoring force that must be overcome by the applied force in order to
deform the object. From Hooke’s law, the restoring force F is proportional to the displacement s provided the
elastic limit is not exceeded. Hence
F r =−ks

Restoring force =−(force constant)(displacement)
The minus sign is required because the restoring force acts in the opposite direction to the displacement. The
greater the value of the force constantk, the greater the restoring force for a given displacement and the greater
the applied force F = ks needed to produce the displacement.

ELASTIC POTENTIAL ENERGY
Because work must be done by an applied force to stretch or compress an object, the object has elastic potential
energy as a result, where

1
PE = ks 2
2
When a deformed elastic object is released, its elastic potential energy turns into kinetic energy or into work
done on something else.

SOLVED PROBLEM 14.1

A force of 5 N compresses a spring by 4 cm. (a) Find the force constant of the spring. (b) Find the elastic
potential energy of the compressed spring.
F 5N
(a) k = = = 125 N/m
s 0.04 m
2
1
2
1
(b) PE = ks = ( )(125 N/m)(0.04 m) = 0.1J
2 2
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