Question

A physical pendulum is positioned so that its center of gravity is above the suspension point. When the pendulum is released it passes the point of stable equilibrium with an angular velocity $\omega$. The period of small oscillations of the pendulum is

• A. $\frac{4\pi }{\omega }$
• B. $\frac{2\pi }{\omega }$
• C. $\frac{\pi }{\omega }$
• D. $\frac{\pi }{2\omega }$

Answer 1

The correct option is A $\frac{4\pi }{\omega }$

Let l = distance between the C.G.C. of the pendulum and its point of suspension o originally the pendulum is in the inverted position and its C.G. is above o. when it falls to the normal (stable) position of equilibrium it's C.G.has fallen by a distance $2l.$ In the equilibrium position, the total energy is equal to $K.E.=\frac{1}{2}\left(I{\omega }^2\right)$ and we have from energy conservation

$\frac{1}{2}I{\omega }^2=mg2l$

or $I=\frac{4mgl}{{\omega }^2}$

Angular frequency of oscillation for a physical pendulum is given by ${\omega }_0^2=\frac{mgl}{l}$

Thus $T=2\pi \times \sqrt{\frac{I}{mgl}}=2\pi \sqrt{\frac{\frac{4mgl}{{\omega }^2}}{mgl}}=\frac{4\pi }{\omega }$