Question

A uniform rod of length \(L\) and mass \(M\) is pivoted at one end and displaced by a small angle \(\theta\) from vertical. The time period of small oscillations is:

• A. \(2\pi\sqrt{\frac{L}{g}}\)
• B. \(2\pi\sqrt{\frac{2L}{3g}}\)
• C. \(2\pi\sqrt{\frac{3L}{2g}}\)
• D. \(2\pi\sqrt{\frac{L}{2g}}\)

Hint:

Always use the physical pendulum formula when pivot is not at centre of mass.

Answer 1

The Correct Option is B

For a physical pendulum, \[ T = 2\pi\sqrt{\frac{I}{Mgh}} \] For a rod about one end: \(I=\frac{1}{3}ML^2,\; h=\frac{L}{2}\) \[ T=2\pi\sqrt{\frac{\frac{1}{3}ML^2}{Mg\frac{L}{2}}} =2\pi\sqrt{\frac{2L}{3g}} \]