A mass M attached to a horizontal spring exec | vedclass

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(C) When the mass $M$ passes through the mean position,its velocity $V$ is maximum,given by $V = \omega_{1} A_{1} = \sqrt{\frac{k}{M}} A_{1}$.When mas

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

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(C) When the mass $M$ passes through the mean position,its velocity $V$ is maximum,given by $V = \omega_{1} A_{1} = \sqrt{\frac{k}{M}} A_{1}$.When mass $m$ is placed on $M$,the momentum is conserved during the collision (as the impulsive force acts internally).$M V = (M + m) V^{\prime}$,where $V^{\prime}$ is the new velocity at the mean position.$V^{\prime} = \frac{M}{M+m} V$.The new amplitude $A_{2}$ is determined by the new angular frequency $\omega_{2} = \sqrt{\frac{k}{M+m}}$.Since $V^{\prime} = \omega_{2} A_{2}$,we have $A_{2} = \frac{V^{\prime}}{\omega_{2}} = \frac{M V}{(M+m) \sqrt{\frac{k}{M+m}}} = \frac{M \sqrt{\frac{k}{M}} A_{1}}{(M+m) \sqrt{\frac{k}{M+m}}} = \frac{M}{\sqrt{M}} \cdot \frac{\sqrt{M+m}}{M+m} A_{1} = \sqrt{\frac{M}{M+m}} A_{1}$.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 $S.H.M.$ 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 of $\left(\frac{A_{1}}{A_{2}}\right)$ is

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