Question

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 _____.

Answer 1

Correct Answer: 7

At \( x = \frac{2A}{3} \),

\[ v = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2} = \omega \sqrt{\frac{5A^2}{9}} = \omega \frac{\sqrt{5}A}{3} \]

New amplitude is \( A' \):

\[ v' = 3v = 3 \left( \omega \frac{\sqrt{5}A}{3} \right) = \omega \sqrt{(A')^2 - \left(\frac{2A}{3}\right)^2} \] \[ \omega \sqrt{5}A = \omega \sqrt{(A')^2 - \left(\frac{2A}{3}\right)^2} \]

Squaring both sides:

\[ 5A^2 = (A')^2 - \left(\frac{2A}{3}\right)^2 \] \[ (A')^2 = 5A^2 + \left(\frac{4A^2}{9}\right) = \frac{45A^2}{9} + \frac{4A^2}{9} = \frac{49A^2}{9} \] \[ A' = \frac{7A}{3} \] \[ \therefore \, n = 7 \]