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A mass ‘M’ is suspended from a light spring An additional mass ‘m’ added displaces the spring further by a distance $' x'$ . Now the combined mass will oscillate in the spring with a period

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T = 2 π \sqrt{\frac{(M + m) x}{mg}}

T = 2 π \sqrt{\frac{M + m}{mgx}}

T = 2 π \sqrt{\frac{mg}{x (M + m)}}

T = 2 π \sqrt{\frac{mgx}{M + m}}

answer is B.

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Detailed Solution

Time period – of spring block system is given by,

$T = 2 π \sqrt{\frac{mass of block}{spring constant}}$

$(M + m) \to$ Mass of block

Spring constant $k = \frac{mg}{x}$

At equilibrium,

$(M + m) g = k (x_{0} + x)$

(or)

$mg = kx$

$T = 2 π \sqrt{\frac{(M + m)}{\frac{mg}{x}}}$

$= 2 π \sqrt{\frac{(M + m) x}{mg}}$

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