In the given figure,a mass M is attached to a | vedclass

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(B) When the mass $M$ is at the extreme position,its velocity is zero. Therefore,the kinetic energy of the system is zero.When the mass $m$ is gently

Vedclass · Accepted Answer

$A \sqrt{\frac{M}{M+m}}$

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(B) When the mass $M$ is at the extreme position,its velocity is zero. Therefore,the kinetic energy of the system is zero.When the mass $m$ is gently placed on $M$ at the extreme position,the total energy of the system remains conserved.The total energy of the system before placing the mass $m$ is $E = \frac{1}{2} k A^2$.After placing the mass $m$,the new mass of the system becomes $(M + m)$. The spring constant $k$ remains the same.Since the mass is placed at the extreme position,the displacement is still $A$. Thus,the potential energy remains $\frac{1}{2} k A^2$.However,the new amplitude $A^{\prime}$ is determined by the new equilibrium position. Since the spring is horizontal and there is no friction,the equilibrium position does not change.Thus,the new amplitude $A^{\prime}$ remains $A$ if placed at the extreme position. However,if the question implies the mass is placed at the equilibrium position (where velocity is maximum),then momentum is conserved:$M v_{max} = (M + m) v^{\prime}_{max}$$M (A \omega) = (M + m) (A^{\prime} \omega^{\prime})$$M A \sqrt{\frac{k}{M}} = (M + m) A^{\prime} \sqrt{\frac{k}{M+m}}$$A \sqrt{M k} = A^{\prime} \sqrt{(M + m) k}$$A^{\prime} = A \sqrt{\frac{M}{M+m}}$

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