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 . The value of is

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Solution

Q.
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 $n A / 3$ . The value of $n$ is ______ . [2024]

Ans.
(7)    $v = ω \sqrt{A^{2} - x^{2}}$

at $x = \frac{2 A}{3}$

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

New amplitude = $A^{'}$

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

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

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