A particle is subjected to two simple harmonic motions as: x 1 =√(7) sin 5 t cm and x 2 =2 √(7) sin (5 t+π/3 ) cm where x is displacement and t is time in seconds. The maximum acceleration of the particle is x × 10 -2 ms -2 . The value of x is :

Physics · Simple Harmonic Motion

JEE Main 2025 — 2 April, Morning Shift — Question 57

A particle is subjected to two simple harmonic motions as: $x_{1}=\sqrt{7} \sin 5 t \mathrm{~cm}$ and

$x_{2}=2 \sqrt{7} \sin \left(5 t+\frac{\pi}{3}\right) \mathrm{cm}$ where $x$ is displacement and $t$ is

time in seconds. The maximum acceleration of the particle is $x \times 10^{-2} \mathrm{~ms}^{-2}$ .

The value of $x$ is :

$\omega=5$

\begin{aligned} A & =\sqrt{(\sqrt{7})^{2}+(2 \sqrt{7})^{2}+2 \times 2 \times 7 \cos \left(\frac{\pi}{3}\right)} & =7 \end{aligned}

a_{\max }=\omega^{2} A=25 \times 7

=175

Answer key and solution verified before publishing.

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