Q 11-13-065JEE MainJEE Main 2024 (31 Jan, Shift 1)Medium
A particle performs simple harmonic motion with amplitude $A$. Its speed is increased to three times at an instant when its displacement is $\dfrac{2A}{3}$. The new amplitude of motion is $\dfrac{nA}{3}$. The value of $n$ is ______.
Numerical value type. Enter your answer.
Answer: 7
At $x = \dfrac{2A}3$: $v = \omega\sqrt{A^2 - \dfrac{4A^2}{9}} = \dfrac{\sqrt5}{3}\omega A$. The new speed is $3v = \sqrt5\,\omega A$ ($\omega$ is unchanged).
$$A'^2 = x^2 + \frac{v'^2}{\omega^2} = \frac{4A^2}{9} + 5A^2 = \frac{49A^2}{9} \Rightarrow A' = \frac{7A}{3}$$
$n = 7$.
Solution by Sreeraj P, M.Sc Physics