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 simple harmonic oscillations with a time period T. If the mass is increased by m, then the time period becomes $(\frac{5}{4} T)$ . The ratio of $\frac{m}{M}$ is

9/16

5/4

25/16

4/5

Solution

9/16

The time period of oscillation is

T = 2 $π \sqrt{\frac{M}{K}}$

M = mass attached to spring

and T₁ = 2 $π \sqrt{\frac{M + m}{K}}$

⇒ $\frac{T_{1}}{T} = \sqrt{\frac{M + m}{M}}$

⇒ $\frac{5}{4} = {(1 + \frac{m}{M})}^{\frac{1}{2}}$ $(∵ T_{1} = \frac{5}{4} T)$

⇒ $\frac{m}{M} = \frac{25}{16} -$ 1

⇒ $\frac{m}{M} = \frac{9}{16}$

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