Question

A 0.497 kg body undergoes simple harmonic motion of amplitude 7.63 cm and period 0.200 s. (a) What is the magnitude of the maximum force acting on it

Answer 1

ω = (m / k)^1/2 frequency of vibration

x = A sin ω t displacement of body at time t

P = 1 / f = 2 π / ω since ω = 2 π f

ω = 2 π / P = 2 π / .2 = 10 * 3.14 = 31.4 / sec

Since a(max) = A ω^2 maximum acceleration

a(max) = .0763 * 31.4^2 = 75.2 m/s^2

F = m a = .497 * 75.2 = 37.4 N maximum force