Explanation

(c) When a mass $m$ is placed on mass$M$, the new system is of mass$= (M + m)$, attached to the spring. New time period of oscillation,
$T^{'} = 2\pi\sqrt{\frac{M + m}{k}}$
$T = 2\pi\sqrt{\frac{M}{k}}$
Let $v =$velocity of the mass $M$ while passing through the mean position.
$v^{'} =$Velocity of the mass$(M + m)$, while passing through the mean position.
According to law of conservation of linear momentum $Mv = (M + m)v^{'}$
At mean position, $v = A\ \omega$ and $v^{'} = A^{'}\omega^{'}$
$\therefore\ \ \ MA\omega = (m + m)A^{'}\omega^{\ }$
or $A^{'} = \left( \frac{M}{M + m} \right)\frac{\omega}{\omega},A = \frac{M}{M + m} \times \frac{T^{'}}{T} \times A$
$= \left( \frac{M}{M + m} \right) \times \sqrt{\frac{M + m}{M}} \times A$
$= A\sqrt{\frac{M}{M + m}}$