A simple harmonic oscillator has a frequency of 2.05 Hz. What will be its amplitude of oscillation if it's started from its equillibrium position with a velocity of 1.9 m/s? Express your answer in m to three significant digits.

Answer 1

x = A sin 2 pi f t

x = A sin 12.9 t
v = dx/dt = 12.9 A cos 12.9 t
v = 1.9 = 12.9 A cos 0 = 12.9 A

A = 1.90/12.9 = .147 meter