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 \(m_2\).
Step 1: Underlying 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: 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: Conclusion:
\[
\boxed{d\sqrt{\frac{m_1}{m_1 + m_2}}}
\]
