In the arrangement shown, all surfaces are smooth and spring is ideal. If block of mass m is given slight displacement, calculate the time period of oscillation of the system.

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$2 π \sqrt{\frac{M m}{(M + m) k}}$

$2 π \sqrt{\frac{M + m}{k M m}}$

$2 π \sqrt{\frac{M - m}{k (M + m)}}$

$2 π \sqrt{\frac{k M m}{(M + m)}}$

answer is B.

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Detailed Solution

$M \frac{d^{2} y}{d t^{2}} = - k (x + y) = - k (\frac{M}{m} y + y) [∵ y = \frac{m}{M} x \Rightarrow x = \frac{M}{m} y] = - k (\frac{M + m}{m}) y$

or $\frac{d^{2} y}{d t^{2}} = - k (\frac{M + m}{M m}) y$

Comparing with, $a = - ω^{2} y$ , we have

$ω = \sqrt{\frac{k (M + m)}{M m}} \Rightarrow T = \frac{2 π}{ω} = 2 π \sqrt{\frac{M m}{(M + m) k}}$

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