In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillation will be :

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

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

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

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

answer is B.

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Detailed Solution

Velocity at mean position is $= A ω$
Conserving momentum $M A ω_{0} = (M + m) V^{1}$

$\begin{array}{l} \Rightarrow V^{1} = \frac{M A ω_{0}}{M + m} \\ Also, New veloctiy at mean position = V^{1} = (A^{'}) ω^{1} = (A^{'}) \sqrt{\frac{K}{M + m}} \\ \Rightarrow \frac{M A ω_{0}}{M + m} = (A^{'}) \sqrt{\frac{K}{M + m}} [here, ω_{0} = \sqrt{\frac{K}{M}}] \\ \Rightarrow A^{'} = \frac{M A \sqrt{\frac{K}{M}}}{M + m} \times \sqrt{\frac{M + m}{K}} = A \sqrt{\frac{M}{(M + m)}} \end{array}$

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