Oscillations

JEE Main level, numerical answer, medium

17 of 194

Question 17 of 194

A particle performs simple harmonic motion with amplitude AA. Its speed is increased to three times at an instant when its displacement is 2A3\dfrac{2A}{3}. The new amplitude of motion is nA3\dfrac{nA}{3}. The value of nn is _____.

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Answer: 7

Speed in SHM in terms of displacement is given by, v=ωA2−x2v=\omega \sqrt{{A}^{2}-{x}^{2}}.
at x=2A3x=\dfrac{2A}{3}, we get
v=ωA2−(2A3)2=5Aω3v=\omega \sqrt{{A}^{2}-{\left(\dfrac{2A}{3}\right)}^{2}}=\dfrac{\sqrt{5}A\omega }{3}
Let the new amplitude be=A′=A'
Then,
v′=3v⇒ω(A′)2−(2A3)2=5Aωv'=3v \Rightarrow \omega \sqrt{{\left(A'\right)}^{2}-{\left(\dfrac{2A}{3}\right)}^{2}}=\sqrt{5}A\omega
⇒A′=73A\Rightarrow A'=\dfrac{7}{3}A
Therefore, n=7n=7.

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