A light spring is suspended with mass $m_1$ at its lower end and its upper end fixed to a rigid support. The…

A light spring is suspended with mass $m_1$ at its lower end and its upper end fixed to a rigid support. The mass is pulled down a short distance and then released. The period of oscillation is $\mathrm{T}$ second. When a mass $\mathrm{m}_2$ is added to $m_1$ and the system is made to oscillate, the period is found to be $\frac{3}{2} \mathrm{~T}$. The ratio $\mathrm{m}_1: \mathrm{m}_2$ is
  1. $2:3$
  2. $3:4$
  3. $4:5$
  4. $5:6$

Solution

The time period is $T \propto \sqrt{\mathrm{m}}$ $\begin{array}{ll} \therefore & \frac{\mathrm{T}}{3}=\frac{\sqrt{\mathrm{m}_1}}{\sqrt{\mathrm{m}_1+\mathrm{m}_2}} \\ \therefore & \frac{2}{3}=\frac{\sqrt{\mathrm{m}_1}}{\sqrt{\mathrm{m}_1+\mathrm{m}_2}} \\ \therefore \quad & \frac{4}{9}=\frac{\mathrm{m}_1}{\mathrm{~m}_1+\mathrm{m}_2} \\ \therefore \quad & 4 \mathrm{~m}_1+4 \mathrm{~m}_2=9 \mathrm{~m}_1 \\ \therefore \quad & 5 \mathrm{~m}_1=4 \mathrm{~m}_2 \\ \therefore \quad & \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{4}{5} \end{array}$ /

Asked in: MHT CET 2023 (14 May Shift 2)

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