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Simple Harmonic Motion

JEE Main 2025 (Online) 2nd April Morning Shift

MCQ+4 / -12025

A particle is subjected to two simple harmonic motions as :

\(x_1=\sqrt{7} \sin 5 \mathrm{tcm}\)

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 :

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