A mass M,attached to a horizontal spring,exec | vedclass

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(D) At the mean position,the potential energy is zero and the kinetic energy is maximum. Since no external horizontal force acts on the system during

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$\left( \frac{M + m}{M} \right)^{\frac{1}{2}}$

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(D) At the mean position,the potential energy is zero and the kinetic energy is maximum. Since no external horizontal force acts on the system during the placement of mass $m$,linear momentum is conserved.Let $v_1$ be the velocity of mass $M$ at the mean position,and $v_2$ be the velocity of the combined mass $(M+m)$ immediately after placing $m$.Conservation of momentum: $M v_1 = (M + m) v_2$.The maximum velocity in $S.H.M.$ is given by $v_{max} = A \omega = A \sqrt{\frac{k}{m_{eff}}}$.For the first case: $v_1 = A_1 \sqrt{\frac{k}{M}}$.For the second case: $v_2 = A_2 \sqrt{\frac{k}{M+m}}$.Substituting these into the momentum equation:$M \left( A_1 \sqrt{\frac{k}{M}} \right) = (M + m) \left( A_2 \sqrt{\frac{k}{M+m}} \right)$.$A_1 \sqrt{M k} = A_2 \sqrt{(M+m) k}$.$A_1 \sqrt{M} = A_2 \sqrt{M+m}$.Therefore,$\frac{A_1}{A_2} = \sqrt{\frac{M+m}{M}} = \left( \frac{M+m}{M} \right)^{\frac{1}{2}}$.

$A$ mass $M$,attached to a horizontal spring,executes $S.H.M.$ with amplitude $A_1$. When the mass $M$ passes through its mean position,a smaller mass $m$ is placed over it and both of them move together with amplitude $A_2$. The ratio of $\frac{A_1}{A_2}$ is

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