Question

A particle performing $S.H.M$ is found at its equilibrium at $t=1s$ and it is found to have a speed of $0.25m/s$ at $t=2s$.If the period of oscillation is $6s$.Calculate amplitude of oscillation

• A. $\frac{3}{2\pi }m$
• B. $\frac{3}{4\pi }m$
• C. $\frac{6}{\pi }m$
• D. $\frac{3}{8\pi }m$

Answer 1

The correct option is A $\frac{3}{2\pi }m$

Given,

$\omega =\frac{2\pi }{6}=\frac{\pi }{3}$

Let the displacement of particle be,

$y=A\sin (\omega t+\varphi )$

At $t=1,y=0$

$\frac{\pi t}{3}+\varphi =\pi$

$\varphi =\frac{2\pi }{3}$

Velocity at $t=1,0.25m/s$

$v=A\omega \cos (\frac{2\pi }{3}+\frac{2\pi }{3})=\frac{A\omega }{2}$

$0.25=\frac{\pi A}{6}$

$A=\frac{3}{2\pi }$

Option $\text{A}$ is the correct answer