Question

A particle executes simple harmonic motion of period 16 s. Two seconds later after it passes through the center of oscillation its velocity is found to be 2 m/s. Find the amplitude.

Answer 1

$\omega =\frac{2\pi }{T}=\frac{2\pi }{16}=\frac{\pi }{8}$ rad/s
If at $t=0,$ particle passes through its mean position $(x=A\sin \omega t)$ with maximum speed its $v-t$ equation can be written as
$v=\omega A\cos \omega t$
Substituting the given values, we have
$2=\left(\frac{\pi }{8}\right)A\cos \left(\frac{\pi }{8}\right)\left(2\right)$
$\therefore$ $A=\frac{16\sqrt{2}}{\pi }$ m$=7.2m$