Page 184 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 14] SIMPLE HARMONIC MOTION 169

SOLVED PROBLEM 14.5
An object of unknown mass is suspended from a spring, which stretches by 10 cm as a result. If the
system is set in oscillation, what will its frequency be?
The force F that causes the spring to stretch by s = 10 cm = 0.1 m is the weight mg of the unknown mass, so
the force constant of the spring is
F mg
k = = = 10mg m −1
s 0.1m
The period of oscillation of the system is therefore

m m 1
T = 2π = 2π = 2π = 0.635 s
2
−1
k 10mg m −1 (10 m )(9.8 m/s )
and the frequency is f = 1/T = 1.58 Hz.
SOLVED PROBLEM 14.6
A spring whose force constant is 12 lb/ft oscillates up and down with a period of 0.5 s when a wrench is
suspended from it. How much does the wrench weigh?
Since m = w/g,

m w
T = 2π = 2π
k gk
2
gkT 2 (32 ft/s )(12 lb/ft)(0.5s) 2
w = = = 2.4lb
4π 2 4π 2
DISPLACEMENT,VELOCITY, AND ACCELERATION
If t = 0 when a body undergoing simple harmonic motion is in its equilibrium position of s = 0 and is moving
in the direction of increasing s, then at any time t thereafter its displacement is
s = A sin 2πft

Often this formula is written

s = A sin ωt
where ω = 2πf is the angular frequency of the motion in radians per second. (If instead the body is at s =+A
when t = 0, then s = A cos 2πft = A cos ωt.) Figure 14-2 is a graph of s versus t.

Fig. 14-2