Page 183 - Schaum's Outline of Theory and Problems of Applied Physics

168 SIMPLE HARMONIC MOTION [CHAP. 14

KE = 0 Frictionless
m
PE = 0 surface Equilibrium position

s
KE = 0
1
PE = ks 2
2 Mass is pulled out a distance s and
released
v
KE = 1 mv 2
2
PE = 0 At s = 0 all the energy is kinetic

s
KE = 0
1 Spring is compressed and all the
PE = ks 2
2 energy is potential

v 1
KE = mv 2
2
The mass is at s = 0 and moves in
PE = 0 the opposite direction

Fig. 14-1
In the case of a body of mass m attached to a spring of force constant k, F r =−ks = ma, and so −s/a = m/k.
Hence

m
T = 2π stretched spring
k
The frequencyf of a body undergoing simple harmonic motion is the number of cycles per second it executes,
so that
1
f =
T
1
Frequency =
period
The unit of frequency is the hertz (Hz), where 1 Hz = 1 cycle/s.

SOLVED PROBLEM 14.4
A 100-g object is suspended from a spring whose force constant is 50 N/m. (a) By how much does the
spring stretch? (b) What is the period of oscillation of the system? (c) What is its frequency?
2
(a) Here F = mg = (0.1 kg)(9.8 m/s ) = 0.98 N, and so the spring stretches by
F 0.98 N
s = = = 0.0196 m = 1.96 cm
k 50 N/m

m 0.1kg
(b) T = 2π = 2π = 0.281 s
k 50 N/m
1 1
(c) f = = = 3.56 s −1 = 3.56 Hz
T 0.281 s