MHTCET 16 Apr 2026 Shift 2 Paper

Concept:At the mean position, the velocity is maximum and momentum is conserved when the second mass is gently placed.Explanation:For the initial motion, the angular frequency is$$\omega = \sqrt{\frac{k}{m_{1}}}$$The maximum speed at the mean position is$$v_{\max} = \omega A = A\sqrt{\frac{k}{m_{1}}}$$When $m_{2}$ is placed on $m_{1}$ at the mean position, the spring force is zero, so horizontal momentum is conserved:$$m_{1}v_{\max} = (m_{1}+m_{2})v'$$Thus the common speed becomes$$v' = \frac{m_{1}}{m_{1}+m_{2}}v_{\max}$$The new angular frequency of the combined system is$$\omega' = \sqrt{\frac{k}{m_{1}+m_{2}}}$$Since the combined mass starts from the mean position with speed $v'$, the new amplitude is$$A_{1} = \frac{v'}{\omega'}$$Substituting the values gives$$A_{1} = \frac{\frac{m_{1}}{m_{1}+m_{2}}A\sqrt{\frac{k}{m_{1}}}}{\sqrt{\frac{k}{m_{1}+m_{2}}}} = A\sqrt{\frac{m_{1}}{m_{1}+m_{2}}}$$Therefore,$$\frac{A}{A_{1}} = \sqrt{\frac{m_{1}+m_{2}}{m_{1}}}$$Answer:$$\frac{A}{A_{1}} = \left[\frac{m_{1}+m_{2}}{m_{1}}\right]^{\frac{1}{2}}$$So, the correct option is A.