JEE Main 2022 — Oscillations Question with Solution

At mean position, the velocity of the particle in S.H.M. is maximum, so kinetic energy is $KE=\frac{1}{2}k{A}^2$, where, $k$ is spring constant.

We also know that, kinetic energy is related to momentum as $KE=\frac{p^2}{2m}$.

Now, from both equations, we have 

$\frac{1}{2}k{A}^2=\frac{p^2}{2\mathrm{m}}$     (Here, momentum is conserved)

$\Rightarrow {\left(\frac{A_1}{ A_2}\right)}^2=\frac{m_2}{ m_1}=\frac{1024}{900}$

$\Rightarrow \frac{A_1}{{ A}_2}=\frac{32}{30}=\frac{16}{15}=\frac{16}{16-1}$

$\therefore \alpha =16$