Q 11-13-156JEE MainJEE Main 2025 (2 Apr, Shift 1)Medium
A particle is subjected to two simple harmonic motions as:
$x_1 = \sqrt7\sin 5t\ \text{cm}$ and $x_2 = 2\sqrt7\sin\left(5t + \dfrac{\pi}{3}\right)\ \text{cm}$
where $x$ is displacement and $t$ is time in seconds. The maximum acceleration of the particle is $x \times 10^{-2}\ \text{m s}^{-2}$. The value of $x$ is:
Answer: (A) $175$
Both motions have the same $\omega = 5\ \text{rad/s}$, so they add to a single SHM with amplitude
$$A^2 = A_1^2 + A_2^2 + 2A_1A_2\cos\frac{\pi}{3} = 7 + 28 + 2(\sqrt7)(2\sqrt7)\left(\tfrac12\right) = 49 \Rightarrow A = 7\ \text{cm}$$
Maximum acceleration:
$$a_{\max} = \omega^2 A = 25 \times 7 = 175\ \text{cm/s}^2 = 175\times10^{-2}\ \text{m/s}^2$$
So $x = 175$.
Solution by Sreeraj P, M.Sc Physics