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

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:

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