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

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:

$\alpha = 12$