### Oscillations Question 7

#### Question 7 - 2024 (31 Jan Shift 1)

A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is $\frac{2 A}{3}$. The new amplitude of motion is $\frac{nA}{3}$. The value of $n$ is

## Show Answer

#### Answer: (7)

#### Solution:

$v=\omega \sqrt{A^{2}-x^{2}}$

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

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

New amplitude $=A^{\prime}$

$v^{\prime}=3 v=\sqrt{5} A \omega=\omega \sqrt{\left(A^{\prime}\right)^{2}-\left(\frac{2 A}{3}\right)^{2}}$

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