JEE Main 2023 — Oscillations Question with Solution

The position of a particle performing an SHM is given by:

$x=A \sin (\omega t+\varphi )$

Since at $t=0 \mathrm{s}$, the particle is at $x=0$, therefore the initial phase $\varphi =0$.

At time $t=2 \mathrm{s}$, the particle is at $x=\frac{A}{2}$.

$$\begin{gathered} \therefore \frac{A}{2}=A \sin \left(2\omega \right) \\ \Rightarrow \omega =\frac{\mathrm{\pi }}{12} {\mathrm{s}}^{-1} \end{gathered}$$

Therefore, time period

$T=\frac{2\mathrm{\pi }}{\omega }=24 \mathrm{s}$

The total time taken from $x=0$ to $x=A$ is $\frac{T}{4}=6 \mathrm{s}$.

Since the time taken from $x=0$ to $x=\frac{A}{2}$ is $2 \mathrm{s}$, therefore the time taken from $x=\frac{A}{2}$ to $x=A$ is $4 \mathrm{s}$.