Question

A mass $m$ is attached to two springs as shown in the figure. The spring constants of the two springs are $K_1$ and $K_2$. For the frictionless surface, the time period of oscillation of mass $m$ is:

• A. \(2 \pi \sqrt{\frac{m}{K_1-K_2}}\)
• B. \(\frac{1}{2 \pi} \sqrt{\frac{K_1-K_2}{m}}\)
• C. \(\frac{1}{2 \pi} \sqrt{\frac{K_1+K_2}{m}}\)
• D. \(2 \pi \sqrt{\frac{m}{K_1+K_2}}\)

Answer 1

The Correct Option is D

To determine the time period of oscillation of the mass \( m \) attached to two springs on a frictionless surface, we need to consider the spring constants \( K_1 \) and \( K_2 \) of the two springs.

When a mass is attached to two springs in series, the equivalent spring constant \( K_{\text{eq}} \) is given by the formula:

K_{\text{eq}} = \frac{K_1 K_2}{K_1 + K_2}

However, in this setup, the springs are arranged parallel to each other with respect to the mass. Thus, the effective spring constant \( K_{\text{eff}} \) is simply the sum of the individual spring constants:

K_{\text{eff}} = K_1 + K_2

The time period \( T \) of oscillation for a mass-spring system on a frictionless surface is given by:

T = 2 \pi \sqrt{\frac{m}{K_{\text{eff}}}}

Substituting the equivalent spring constant:

T = 2 \pi \sqrt{\frac{m}{K_1 + K_2}}

This matches the correct answer: \(2 \pi \sqrt{\frac{m}{K_1+K_2}}\).

Conclusion: The time period of oscillation of the mass \( m \) is given by \( 2 \pi \sqrt{\frac{m}{K_1+K_2}} \), reflecting that the springs contribute together when attached to the mass in parallel.