Question

A point moves in a straight line with simple harmonic motion. When it is 500mm to the right of the mean position and moving to the left, the velocity is 3m/s. When it is 1 meter to the left of the mean position and moving to the right, the velocity is 2m/s. Determine the amplitude.

Answer 1

$tan\phi = \frac{v(0)}{\omega x(0)} = \frac{3}{2\times0.5}$

$\phi = 71.56^0$

Amplitude :

$x_m = \frac{x(0)}{cos\phi} = \frac{1}{cos(71.56)} =3.16 m = 316 cm$