Question:
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 $T$ second. When a mass $m_2$ is added to $m_1$ and the system is made to oscillate, the period is found to be $\frac{3}{2}T$. The ratio $m_1 : m_2$ is
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 $T$ second. When a mass $m_2$ is added to $m_1$ and the system is made to oscillate, the period is found to be $\frac{3}{2}T$. The ratio $m_1 : m_2$ is
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Since $T \propto \sqrt{m}$, doubling the mass increases the period by a factor of $\sqrt{2}$.
Updated On: Apr 28, 2026
- 2 : 3
- 3 : 4
- 4 : 5
- 5 : 6
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The Correct Option is A
Solution and Explanation
Step 1: Recall the formula for the time period of a spring-mass system. \[ T = 2\pi \sqrt{\frac{m}{k \]
Step 2: Write the expressions for the initial period ($T$) and the new period ($T'$) after adding mass $m_2$. \[ T = 2\pi \sqrt{\frac{m_1}{k \] \[ T' = 2\pi \sqrt{\frac{m_1 + m_2}{k = \frac{3}{2}T \]
Step 3: Divide the second equation by the first to find the ratio of masses. \[ \frac{T'}{T} = \sqrt{\frac{m_1 + m_2}{m_1 = \frac{3}{2} \]
Step 4: Square both sides and solve for the ratio $m_1 : m_2$. \[ \frac{m_1 + m_2}{m_1} = \frac{9}{4} \] \[ 4m_1 + 4m_2 = 9m_1 \] \[ 5m_1 = 4m_2 \implies \frac{m_1}{m_2} = \frac{4}{5} \]
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