Question

System is released after slightly stretching it. Find angular frequency of its oscillations:

• A. \(5\)
• B. \(10\sqrt{5}\)
• C. \(2\sqrt{5}\)
• D. \(5\sqrt{5}\)

Hint:

For two masses connected by a spring on a smooth surface, always use {reduced mass} to find the angular frequency.

Answer 1

The Correct Option is D

To find the angular frequency of the oscillations for this system, we can use the formula for the angular frequency of a mass-spring system:

\(\omega = \sqrt{\frac{k}{m_{\text{eff}}}}\)

where:

• \(k\) is the spring constant.
• \(m_{\text{eff}}\) is the effective mass of the system.

Given values from the diagram:

• Spring constant, \(k = 150 \, \text{N/m}\)
• Masses of the blocks: \(m_1 = 2 \, \text{kg}\), \(m_2 = 3 \, \text{kg}\)

Because both blocks move together, they contribute to the system's effective mass. Therefore, the effective mass \(m_{\text{eff}}\) is the reduced mass given by:

\(m_{\text{eff}} = \frac{m_1 \cdot m_2}{m_1 + m_2}\)

Substituting the known values:

\(m_{\text{eff}} = \frac{2 \times 3}{2 + 3} = \frac{6}{5} = 1.2 \, \text{kg}\)

Now, substitute \(k\) and \(m_{\text{eff}}\) into the angular frequency formula:

\(\omega = \sqrt{\frac{150}{1.2}}\)

\(= \sqrt{125}\)

\(= 5\sqrt{5}\)\)

This means the angular frequency of the system is \(5\sqrt{5}\).

Conclusion: The correct answer is \(\boxed{5\sqrt{5}}\).