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Q.

A mass ‘M’ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period ‘T', if the mass is increased by ‘m’, time period becomes $\frac{5 T}{3}$ , then the ratio of (m / M ) is

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a

$\frac{3}{2}$

b

$\frac{25}{9}$

c

$\frac{16}{9}$

d

$\frac{5}{3}$

answer is C.

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Detailed Solution

$\begin{array}{l} T = 2 π \sqrt{\frac{M}{k}} \Rightarrow T^{1} = 2 π \sqrt{\frac{M + m}{k}} \\ \frac{T^{1}}{T} = \sqrt{\frac{M + m}{M}} \\ \Rightarrow \frac{5}{3} = \sqrt{\frac{M + m}{M}} \\ 25 M = 9 M + 9 m \Rightarrow 16 M = 9 m \\ \frac{16}{9} = \frac{m}{M} \end{array}$

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