Question

A particle performing S.H.M. starts from equilibrium position and its time period is $16$ second. After $2$ seconds its velocity is $\pi m/s$. Amplitude of oscillation is $(\cos {45}^{\circ }=\frac{1}{\sqrt{2}})$

• A. $2\sqrt{2}m$
• B. $4\sqrt{2}m$
• C. $6\sqrt{2}m$
• D. $8\sqrt{2}m$

Answer 1

The correct option is D $8\sqrt{2}m$

let $x=asin\omega t$

on differentiating both side $v=A\omega cos\omega t$

$\pi =A\times \frac{2\pi }{T}cos\frac{2\pi }{T}t$

on putting T=16 sec and t=2 sec

$8=Acos45$

$A=8\sqrt{2}$