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:

• A. \(\left( \frac{1}{m_1} + \frac{1}{m_2} \right) k^2\)
• B. \(\sqrt{\left( \frac{1}{m_1} + \frac{1}{m_2} \right) k^2}\)
• C. \(\sqrt{\left( \frac{1}{m_1} + \frac{1}{m_2} \right)}\)
• D. \(\sqrt{\left( \frac{1}{m_1} + \frac{1}{m_2} \right) k}\)

Hint:

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.

Answer 1

The Correct Option is D

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: \[ \omega = \sqrt{k \left( \frac{1}{m_1} + \frac{1}{m_2} \right)} \]