A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is 2A/3. The new amplitude of motion is nA/3. The value of n is ___________. Numerical answer type. Work it out, then open the answer.

Answer: 7. At x = 2A/3: v = ω√(A² - 4A²/9) = √(5)/3ω A. New speed v' = 3v = √(5) ω A with the same ω. A'² = 4A²/9 + 5A² = (49A²)/(9) ⇒ A' = 7A/3. So n = 7.

A particle performs simple harmonic motion with amplitude A. Its… | JEE Main 2024 (31 Jan, Shift 1)

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Velocity & Acceleration in SHM

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 ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 7

At $x = \frac{2A}{3}$ : $v = \omega\sqrt{A^2 - \frac{4A^2}{9}} = \frac{\sqrt{5}}{3}\omega A$ .
New speed $v' = 3v = \sqrt{5}\,\omega A$ with the same $\omega$ .
$A'^2 = \frac{4A^2}{9} + 5A^2 = \frac{49A^2}{9} \Rightarrow A' = \frac{7A}{3}$ .
So $n = 7$ .

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