A mass 0 . 9kg , attached to a horizontal spring, executes SHM with...

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

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

Now, from both equations, we have

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

$⇒ {\left(\dfrac{{A}_{1}}{{A}_{2}}\right)}^{2} = \dfrac{{m}_{2}}{{m}_{1}} = \dfrac{1024}{900}$

$⇒ \dfrac{{A}_{1}}{{A}_{2}} = \dfrac{32}{30} = \dfrac{16}{15} = \dfrac{16}{16 - 1}$

$∴ \alpha = 16$