Question

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

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

Answer 1

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

Given that, $T=6\text{s}\Rightarrow \omega =\frac{2\pi }{6}\text{rad/s}$
The general equation of $\text{SHM}$ is
$x=A\sin (\omega t+\varphi )$
$\because$ At $t=1\text{s};x=0$
$\Rightarrow 0=A\sin \left[\right(\omega \times 1)+\varphi ]$ $\Rightarrow \varphi =2n\pi -\omega$ ($n$ is the integer)
Now, $v=\frac{dx}{dt}=A\omega \cos (\omega t+\varphi )$
$\because$ at $t=2\text{s};v=\frac{1}{4}\text{m/s}$
$\Rightarrow \frac{1}{4}=A\omega \cos \left[\right(\omega \times 2)+2n\pi -\omega ]$
$\Rightarrow \frac{1}{4}=A\omega \cos (2n\pi +\omega )=A\omega \cos \left(\omega \right)$
By putting values
$\Rightarrow \frac{1}{4}=A\times \frac{2\pi }{6}\cos \frac{2\pi }{6}$
$\Rightarrow A=\frac{3}{2\pi }\text{m}$
Hence, option $\left(a\right)$ is the correct answer.