Question

A particle performing $SHM$ is found at its equilibrium at $t=1sec$ and it is found to have a speed of $0.25m/s$ at $t=2s$. If the period of oscillation is $6sec$, 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

Answer: (A)

$\frac{3}{2\pi }m$

Since a SHM can be represented by $x=A\sin (\omega t+\varphi )$
$\Rightarrow x=A\sin (\left(\frac{2\pi }{T}\right)t+\varphi )$ ; $T=6s$
so the equation becomes
$x=A\sin (\frac{\pi t}{3}+\varphi )$
We are given that at $t=1s,x=0$; so we have
$0=A\sin (\frac{\pi }{3}+\varphi )\Rightarrow (\frac{\pi }{3}+\varphi )=0\Rightarrow \varphi =-\frac{\pi }{3}\Rightarrow x=A\sin (\frac{\pi t}{3}-\frac{\pi }{3})\Rightarrow v=\frac{dx}{dt}=\frac{d}{dt}\left(A\sin (\frac{\pi t}{3}-\frac{\pi }{3})\right)=\frac{\pi A}{3}\cos (\frac{\pi t}{3}-\frac{\pi }{3})$
Now we are given that at $t=2s,\left|v\right|=0.25m/s$, so we have
$0.25=\left|\frac{\pi A}{3}\cos (\frac{2\pi }{3}-\frac{\pi }{3})\right|\Rightarrow 0.25=\left|\frac{\pi A}{3}\cos \left(\frac{\pi }{3}\right)\right|\Rightarrow 0.25=\frac{\pi A}{6}\Rightarrow A=\frac{3}{2\pi }$