Question

A uniform stick of length l and mass m pivoted at its top end is oscillating with an angular frequency ω_r . Assuming small oscillations, the ratio ω_r/ω_s , where ω_s is the angular frequency of a simple pendulum of the same length, will be

• A. \(\sqrt3\)
• B. \(\sqrt{\frac{3}{2}}\)
• C. \(\sqrt2\)
• D. \(\frac{1}{\sqrt3}\)

Answer 1

The Correct Option is B

To solve this problem, we need to compare the angular frequency \( \omega_r \) of a physical pendulum (the stick) to \( \omega_s \), the angular frequency of a simple pendulum.

1. Angular Frequency of a Physical Pendulum (Stick):
   • The stick is pivoted at one end, which makes it a physical pendulum.
   • For small oscillations, the angular frequency \( \omega_r \) is given by: \(\omega_r = \sqrt{\frac{mgd}{I}}\) , where \(d\) is the distance from the pivot to the center of mass and \(I\) is the moment of inertia about the pivot.
   • For a uniform stick, \(d = \frac{l}{2}\) and the moment of inertia \(I = \frac{1}{3}ml^2\) .
   • Substituting these values gives: \(\omega_r = \sqrt{\frac{mg\left(\frac{l}{2}\right)}{\frac{1}{3}ml^2}} = \sqrt{\frac{3g}{2l}}\) .
2. Angular Frequency of a Simple Pendulum:
   • For a simple pendulum of length \(l\) , the angular frequency \(\omega_s\) is: \(\omega_s = \sqrt{\frac{g}{l}}\) .
3. Ratio of Angular Frequencies:
   • We are required to find \(\frac{\omega_r}{\omega_s}\) :
   • Using the expressions derived: \(\frac{\omega_r}{\omega_s} = \frac{\sqrt{\frac{3g}{2l}}}{\sqrt{\frac{g}{l}}} = \sqrt{\frac{3g}{2l} \times \frac{l}{g}} = \sqrt{\frac{3}{2}}\) .
   • The ratio of the angular frequency of a physical pendulum to that of a simple pendulum is \(\sqrt{\frac{3}{2}}\) .
   • Therefore, the correct answer is \(\sqrt{\frac{3}{2}}\) .