Question

The time period of simple harmonic motion of mass \(M\) in the given figure is \(\pi \sqrt{\frac{\alpha M}{5K}}\), where the value of \(\alpha\) is ______.

Answer 1

Correct Answer: 12

Given system parameters:

The equivalent spring constant for the system is calculated as:

\[ k_{\text{eq}} = \frac{2k \cdot k}{2k + k} + k = \frac{5k}{3} \]

The angular frequency of oscillation (\( \omega \)) is given by:

\[ \omega = \sqrt{\frac{k_{\text{eq}}}{m}} \]

Substituting the value of \( k_{\text{eq}} \):

\[ \omega = \sqrt{\frac{\frac{5k}{3}}{m}} = \sqrt{\frac{5k}{3m}} \]

The period of oscillation (\( \tau \)) is:

\[ \tau = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{\frac{5k}{3}}} = 2\pi \sqrt{\frac{3m}{5k}} \]

Simplifying:

\[ \tau = \pi \sqrt{\frac{12m}{5k}} \]

Thus, comparing with the given expression:

\[ T = \pi \sqrt{\frac{\alpha M}{5K}} \]

we find:

\[ \alpha = 12 \]