The time period of mass M when displaced from its equilibrium position and then released for the system as shown in figure is:

2 π \sqrt{\frac{M}{k}}

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2 π \sqrt{\frac{M}{2 k}}

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2 π \sqrt{\frac{M}{4 k}}

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2 π \sqrt{\frac{2 M}{k}}

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Solution

The correct option is D $2 π \sqrt{\frac{M}{4 k}}$
When mass M is suspended from the given system as shown in figure, let l be the length through which mass M moves down before it comes to rest. In this situation, both the spring and string be stretched by length l. Since string is inextensible, so spring is stretched by length $2 l$ . The tension along the string and spring is the same.
In equilibrium, $M g = 2 (k 2 l)$
If mass M is pulled down through small distance x, then
$F = M g - 2 k (2 l + 2 x) = - 4 k x$ .......(i)
$F \propto x$ and -ve sign shows that it is directed towards mean position. Hence, the mass executes simple harmonic motion.
For SHM, $F = - k x$
Comparing (i) and (ii), we get $k = 4 k$
$∴$ Time period, $T = 2 π \sqrt{\frac{M}{4 k}}$

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