Question

Two very small balls, each having mass m, are attached to two massless rods of length I. Now these rods are joined to form V like figure having angle 60°. This assembly is now hinged in a vertical plane so that it can rotate without any friction about a horizontal axis (perpendicular to the plane of figure) as shown in the figure. The period of small oscillation of this assembly is

• A. $2\pi\sqrt{\frac{l}{g}}$
• B. $2\pi\sqrt{\frac{l}{2g}}$
• C. $2\pi\sqrt{\frac{2l}{\sqrt{3}g}}$
• D. $\frac{1}{2\pi}\sqrt{\frac{\sqrt{3}g}{l}}$

Answer 1

Answer: (C)

$2\pi\sqrt{\frac{2l}{g\sqrt{3}}}$

Answer 2

The period of small oscillation of the assembly can be found by:

1. Calculating the moment of inertia $I$ of the system.
2. Determining the change in potential energy $\Delta U$ when the assembly is displaced by a small angle.
3. Using the formula for the period of a physical pendulum $T = 2\pi\sqrt{\frac{I}{mgd}}$, where $d$ is the distance from the pivot to the center of mass.

The moment of inertia $I$ is $2ml^2$. The effective $k$ is $2mgl\cos(30^\circ)$. The angular frequency $\omega$ is $\sqrt{\frac{g\sqrt{3}}{2l}}$. Therefore, the period $T$ is $2\pi\sqrt{\frac{2l}{g\sqrt{3}}}$.