Question

The ends of a rod of length l and mass m are attached to two identical springs as shown in fig. The rod is free to rotate about its centre o. The rod is depressed slightly at end a and released. The time period of the resulting oscillation is

Answer 1

Answer:

$T = \pi \sqrt{\frac{2m}{3k}$

Explanation:

When the rod is pressed down on one side, the other side rises up by the same amount. Let the rod be pressed down by an amount x.

This causes a restoring torque to be formed.

ζ = force x ⊥r distance

$\tau = \frac{F_sl}{2} + \frac{F_sl}{2}$

$\tau = F_sl$

Restoring force

$F_s = -kx$

Now, from the figure, we can see that the rod is displaced by an angle θ.

$tan\theta = \frac{x}{\frac{l}{2}}\\\\tan\theta = \frac{2x}{l}$

as θ is very small, tanθ ≈ θ

x = lθ/2

Substituting the formula for restoring force and value of x in the formula for torque

$\tau = -\frac{k\theta l^2}{2}$

Now, we know that,

Torque = Iα

Moment of inertia for a rod about its centre

$I = \frac{Ml^2}{12}$

Thus, we can say that

$\frac{Ml^2}{12} \alpha = -\frac{k \theta l^2}{2}$

$\alpha = -\frac{6k \theta}{M} = -\omega^2 \theta$

$\omega = \sqrt{\frac{6k}{M}} \\\\\omega = \frac{2 \pi}{T}\\\\T = \pi \sqrt{\frac{2m}{3k}}$