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(C) At the mean position,the velocity of the mass $M$ is maximum,given by $v_{max} = \omega_1 A_1 = \sqrt{\frac{k}{M}} A_1$. When mass $m$ is placed o

Vedclass · Accepted Answer

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

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(C) At the mean position,the velocity of the mass $M$ is maximum,given by $v_{max} = \omega_1 A_1 = \sqrt{\frac{k}{M}} A_1$. When mass $m$ is placed on $M$,the total mass becomes $(M+m)$. Since the collision is perfectly inelastic and occurs at the mean position,the momentum is conserved. $M v_{max} = (M+m) v'_{max}$. So,$v'_{max} = \frac{M}{M+m} v_{max}$. The new angular frequency is $\omega_2 = \sqrt{\frac{k}{M+m}}$. Since $v'_{max} = \omega_2 A_2$,we have $\frac{M}{M+m} (\omega_1 A_1) = \omega_2 A_2$. Substituting the values: $\frac{M}{M+m} \sqrt{\frac{k}{M}} A_1 = \sqrt{\frac{k}{M+m}} A_2$. $\frac{A_1}{A_2} = \frac{M+m}{M} \sqrt{\frac{k}{M+m}} \sqrt{\frac{M}{k}} = \frac{M+m}{M} \sqrt{\frac{M}{M+m}} = \sqrt{\frac{M+m}{M}} = \left(\frac{M+m}{M}\right)^{\frac{1}{2}}$.

$A$ mass $M$ attached to a horizontal spring executes $S.H.M.$ of 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 $\left(\frac{A_1}{A_2}\right)$ is

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