A particle performs simple harmonic motion… - JEE Main 2024 Physics PYQ

JEE Main 2024 — Physics Waves & Oscillations

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A particle performs simple harmonic motion with amplitude $A$ . Its speed is increased to three times at an instant when its displacement is $\frac{2A}{3}$ . The new amplitude of motion is $\frac{nA}{3}$ . The value of $n$ is _____.

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Held on 31 Jan 2024 · Verified 6 Jul 2026.

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Solution

Speed in SHM in terms of displacement is given by, $v=\omega \sqrt{{A}^{2}-{x}^{2}}$ .

at $x=\frac{2A}{3}$ , we get

$v=\omega \sqrt{{A}^{2}-{(\frac{2A}{3})}^{2}}=\frac{\sqrt{5}A\omega }{3}$

Let the new amplitude be $=A'$

Then,

$v'=3v \Rightarrow \omega \sqrt{{(A')}^{2}-{(\frac{2A}{3})}^{2}}=\sqrt{5}A\omega$

$\Rightarrow A'=\frac{7}{3}A$

Therefore, $n=7$ .

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