A mass ' $\mathrm{m}_1$ ' is suspended from a spring of negligible mass. A spring is pulled slightly in…

A mass ' $\mathrm{m}_1$ ' is suspended from a spring of negligible mass. A spring is pulled slightly in downward direction and released; mass performs S.H.M. of period ' $\mathrm{T}$ ' '. If the mass is increased by ' $\mathrm{m}_2$ ', the time period becomes ' $\mathrm{T}_2$ '. The ratio $\frac{\mathrm{m}_2}{\mathrm{~m}_1}$ is
  1. $\frac{\mathrm{T}_1^2+\mathrm{T}_2^2}{\mathrm{~T}_1^2}$
  2. $\frac{T_1-T_2}{T_1}$
  3. $\frac{\mathrm{T}_2^2-\mathrm{T}_1^2}{\mathrm{~T}_1^2}$
  4. $\frac{\mathrm{T}_1^2-\mathrm{T}_2^2}{\mathrm{~T}_1^2}$

Solution

$\begin{aligned} & \mathrm{T}_1=2 \pi \sqrt{\frac{\mathrm{m}_1}{\mathrm{k}}} \text { and } \mathrm{T}_2=2 \pi \sqrt{\frac{\mathrm{m}_1+\mathrm{m}_2}{\mathrm{k}}} \\ & \therefore \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\sqrt{\frac{\mathrm{m}_1+\mathrm{m}_2}{\mathrm{~m}_1}} \\ & \therefore \frac{\mathrm{T}_2^2}{\mathrm{~T}_1^2}=\frac{\mathrm{m}_1+\mathrm{m}_2}{\mathrm{~m}_1} \\ & \therefore \frac{\mathrm{T}_2^2-\mathrm{T}_1^2}{\mathrm{~T}_1^2}=\frac{\mathrm{m}_2}{\mathrm{~m}_1} \end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 2)

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