Physics · 2025
JEE Main · 2 April 2025, Shift 1 · Q33
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…
A particle is subjected to two simple harmonic motions as :
$$x_1=\sqrt{7} \sin 5 \mathrm{t} \mathrm{~cm}
$$
and $\displaystyle x_2=2 \sqrt{7} \sin \left(5 \mathrm{t}+\frac{\pi}{3}\right) \mathrm{cm}$
where $\displaystyle x$ is displacement and $\displaystyle t$ is time in seconds.
The maximum acceleration of the particle is $\displaystyle x \times 10^{-2} \mathrm{~ms}^{-2}$. The value of $\displaystyle x$ is :
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle 175$
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JEE Main 2025 Physics question, with the answer from NTA’s final answer key. Where our answers come from.