Q132 / 155 · WBJEE 2016 · Physics
A particle vibrating simple harmonically hat an acceleration of $16 \mathrm{cms}^{-2}$ when it is at distance of $4 \mathrm{cm}$ from the mean position, lt time period is
- A$1 \mathrm{s}$
- B$2.572 \mathrm{s}$
- Correct answerC$3.142 \mathrm{s}$
- D$6.028 \mathrm{s}$
Solution
In SHM, the acceleration of vibrating particle is proportional to displacement
$$
\begin{aligned}
|\mathbf{a}| &=\omega^{2} x \\
\text { Here } a=16 \mathrm{cm} / \mathrm{s}^{2} \\
&=16 \times 10^{-2} \mathrm{m} / \mathrm{s}^{2}
\end{aligned}
$$
Displacement, $x=4 \mathrm{cm}=4 \times 10^{-2} \mathrm{m}$
$\therefore \quad 16 \times 10^{-2}=\omega^{2}\left(4 \times 10^{-2}\right)$
$$
\begin{aligned}
\omega^{2} &=4 \\
\omega &=\pm 2 \mathrm{rad} / \mathrm{s} \\
\omega &=\frac{2 \pi}{T} \\
T &=\frac{2 \pi}{\omega}=\frac{2 \pi}{2}=\pi=3.142 \mathrm{s}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.