Question

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

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

Answer 1

The correct option is 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:

$0=A\sin (\frac{\pi }{3}+\varphi )\Rightarrow (\frac{\pi }{3}+\varphi )=0\Rightarrow \varphi =-\frac{\pi }{3}\Rightarrow x=A\sin (\frac{\pi t}{3}-\frac{\pi }{3})\Rightarrow v=\frac{dx}{dt}=\frac{d}{dt}\left(\sin (\frac{\pi t}{3}-\frac{\pi }{3})\right)=\frac{\pi A}{3}\cos (\frac{\pi t}{3}-\frac{\pi }{3})$

Now we are given that at $t=2s,\left|v\right|=0.25m/s$,

So we have:
$0.25=\left|\frac{\pi A}{3}\cos (\frac{2\pi }{3}-\frac{\pi }{3})\right|\Rightarrow 0.25=\left|\frac{\pi A}{3}\cos \left(\frac{\pi }{3}\right)\right|\Rightarrow 0.25=\frac{\pi A}{6}\Rightarrow A=\frac{3}{2\pi }$