Question:

As shown in the figure, two blocks of masses \(m_1\) and \(m_2\) are connected to a spring of force constant \(k\). The blocks are slightly displaced in opposite directions to \(x_1, x_2\) distances and released. If the system executes simple harmonic motion, then the frequency of oscillation of the system (\(\omega\)) is:
twoblocksofmassesm1andm2

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In a system with two masses connected by a spring, the effective mass is given by the reduced mass formula. The frequency of oscillation depends on both the mass terms and the spring constant.
Updated On: Mar 13, 2025
  • \(\left( \frac{1}{m_1} + \frac{1}{m_2} \right) k^2\)
  • \(\sqrt{\left( \frac{1}{m_1} + \frac{1}{m_2} \right) k^2}\)
  • \(\sqrt{\left( \frac{1}{m_1} + \frac{1}{m_2} \right)}\)
  • \(\sqrt{\left( \frac{1}{m_1} + \frac{1}{m_2} \right) k}\)
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The Correct Option is D

Solution and Explanation

Step 1: Define the System's Dynamics For a system of two masses connected by a spring with force constant \( k \), the frequency of oscillation can be determined using the concept of reduced mass.
Step 2: Use the Effective Mass Formula The reduced mass of the system is given by: \[ \mu = \frac{m_1 m_2}{m_1 + m_2} \] The angular frequency of oscillation is given by: \[ \omega = \sqrt{\frac{k}{\mu}} \] Substituting \(\mu = \frac{m_1 m_2}{m_1 + m_2}\) into the equation: \[ \omega = \sqrt{k \left( \frac{1}{m_1} + \frac{1}{m_2} \right)} \]
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