Question:
The two blocks of masses \(m_1\) and \(m_2\) are kept on a smooth horizontal table as shown. Block of mass \(m_1\) but not \(m_2\) is fastened to the spring. If now both the blocks are pushed to the left, so that the spring is compressed at a distance \(d\). The amplitude of oscillation of block of mass \(m_1\) after the system is released, is
The two blocks of masses \(m_1\) and \(m_2\) are kept on a smooth horizontal table as shown. Block of mass \(m_1\) but not \(m_2\) is fastened to the spring. If now both the blocks are pushed to the left, so that the spring is compressed at a distance \(d\). The amplitude of oscillation of block of mass \(m_1\) after the system is released, is
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The amplitude of \(m_1\) is less than \(d\) because some energy is carried away by \(m_2\).
Updated On: Apr 20, 2026
- \(d\sqrt{\frac{m_1}{m_1+m_2}}\)
- \(d\sqrt{\frac{m_2}{m_1+m_2}}\)
- \(d\sqrt{\frac{m_2}{m_1+m_2}}\)
- \(d\sqrt{\frac{2m_1}{m_1+m_2}}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Use energy conservation. Spring potential energy converts into kinetic energy of both blocks. After separation, \(m_1\) oscillates with its share of energy.
Step 2: Detailed Explanation:
Initial potential energy: \(PE = \frac{1}{2}kd^2\). At natural length, both blocks have same velocity \(v\): \[ \frac{1}{2}kd^2 = \frac{1}{2}(m_1 + m_2)v^2 \] Kinetic energy of \(m_1\): \[ KE_{m_1} = \frac{1}{2}m_1v^2 = \frac{1}{2}m_1 \cdot \frac{kd^2}{m_1 + m_2} = \frac{1}{2}kd^2 \cdot \frac{m_1}{m_1 + m_2} \] This becomes maximum potential energy of spring: \[ \frac{1}{2}kA^2 = \frac{1}{2}kd^2 \cdot \frac{m_1}{m_1 + m_2} \] Solve for amplitude: \(A = d\sqrt{\frac{m_1}{m_1 + m_2}}\).
Step 3: Final Answer:
\[ \boxed{d\sqrt{\frac{m_1}{m_1 + m_2}}} \]
Use energy conservation. Spring potential energy converts into kinetic energy of both blocks. After separation, \(m_1\) oscillates with its share of energy.
Step 2: Detailed Explanation:
Initial potential energy: \(PE = \frac{1}{2}kd^2\). At natural length, both blocks have same velocity \(v\): \[ \frac{1}{2}kd^2 = \frac{1}{2}(m_1 + m_2)v^2 \] Kinetic energy of \(m_1\): \[ KE_{m_1} = \frac{1}{2}m_1v^2 = \frac{1}{2}m_1 \cdot \frac{kd^2}{m_1 + m_2} = \frac{1}{2}kd^2 \cdot \frac{m_1}{m_1 + m_2} \] This becomes maximum potential energy of spring: \[ \frac{1}{2}kA^2 = \frac{1}{2}kd^2 \cdot \frac{m_1}{m_1 + m_2} \] Solve for amplitude: \(A = d\sqrt{\frac{m_1}{m_1 + m_2}}\).
Step 3: Final Answer:
\[ \boxed{d\sqrt{\frac{m_1}{m_1 + m_2}}} \]
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