Page 185 - Schaum's Outline of Theory and Problems of Applied Physics

170 SIMPLE HARMONIC MOTION [CHAP. 14

The velocity of the body at the time t is

v = 2πfA cos 2πft = ωA cos ωt

When v is +, the body is moving in the direction of increasing s; when v is −, it is moving in the direction of
decreasing s. In terms of the displacement s, the magnitude of the velocity is

2
v = 2πf A − s 2
The acceleration of the body at time t is

2
2
2
a =−4π f A sin 2πft =−ω A sin ωt
In terms of the displacement s, the acceleration is
2
2
a = 4π f s
SOLVED PROBLEM 14.7

An object is oscillating in simple harmonic motion with an amplitude of 10 cm and a period of 2 s. Find
the magnitudes of its velocity and acceleration when its displacement s from its equilibrium position is
(a)0,(b) +5 cm, and (c) −10 cm.
−1
(a) The frequency of the oscillations is f = 1/T = 0.5s .At s = 0,
−1
2
2
v = 2πf A − s = 2πfA = (2π)(0.5s )(0.10 m) = 0.314 m/s
2
2
a =−4π f s = 0
(b)At s =+5 cm,

−1
2
2
2
v = 2πf A − s = (2π)(0.5s ) (0.10 m) − (0.05 m) = 0.272 m/s
2
−1 2
2
2
2
a =−4π f s =−(4π )(0.5s ) (0.05 m) =−0.493 m/s 2
(c) At s =−10 cm =−A,

2
2
v = 2πf A − s = 0
2
−1 2
2
2
a =−4π f s =−(4π )(0.5s ) (0.10 m) = 0.987 m/s 2
SOLVED PROBLEM 14.8
What is the maximum velocity of a body undergoing simple harmonic motion? At what displacement
does this velocity occur?
√
2
For a body in simple harmonic motion, v = 2πf A − s , which is a maximum when s = 0. Hence
2

2
v max = 2πf A − 0 = 2πfA
This formula gives only the magnitude of v max , not its direction.
SOLVED PROBLEM 14.9
What is the maximum acceleration of a body undergoing simple harmonic motion? At what displacement
does this acceleration occur?