Question

A thin rod of uniform density and length \(2\sqrt{3}\, \text{m}\) is undergoing small oscillations about a pivot point. The time period of oscillation (\(T_m\)) is minimum when the distance of the pivot point from the center-of-mass of the rod is \(x_m\). Which of the following is/are correct? (Assume acceleration due to gravity \(g = 10\, \text{m/s}^2\))

• A. \(x_m = 1\, \text{m}\)
• B. \(x_m = \dfrac{\sqrt{3}}{2}\, \text{m}\)
• C. \(T_m = \dfrac{2\pi}{\sqrt{5}}\, \text{s}\)
• D. \(T_m = \dfrac{2\pi}{\sqrt{5}}\, \text{s}\)

Hint:

For minimum period in physical pendulum, the distance of pivot from center of mass is given by \(x = \sqrt{I_{cm}/m}\).

Answer 1

The Correct Option is A, D

Step 1: Apply the time period formula for a physical pendulum.
\[ T = 2\pi \sqrt{\frac{I_p}{mgx}} \] where \(I_p = I_{cm} + mx^2\), and for a uniform rod, \(I_{cm} = \frac{1}{12}mL^2\). Given \(L = 2\sqrt{3}\, \text{m}\), so \(I_{cm} = m\).

Step 2: Substitute and minimize \(T\).
\[ T = 2\pi \sqrt{\frac{m(x^2 + 1)}{mgx}} = 2\pi \sqrt{\frac{x^2 + 1}{gx}} \] For minimum \(T\), differentiate with respect to \(x\) and set to zero: \[ \frac{d}{dx}\left(\frac{x^2 + 1}{x}\right) = 0 $\Rightarrow$ \frac{2x^2 - (x^2 + 1)}{x^2} = 0 $\Rightarrow$ x^2 = 1 $\Rightarrow$ x = 1 \] But since the rod length is \(2\sqrt{3}\), we use \(x_m = \dfrac{\sqrt{3}}{2}\, \text{m}\).

Step 3: Compute minimum period.
\[ T_m = 2\pi \sqrt{\frac{(x_m^2 + 1)}{g x_m}} = 2\pi \sqrt{\frac{(\frac{3}{4} + 1)}{10 \times \frac{\sqrt{3}}{2}}} = \dfrac{2\pi}{\sqrt{5}}\, \text{s} \]

Step 4: Conclusion.
Hence, \(x_m = \dfrac{\sqrt{3}}{2}\, \text{m}\) and \(T_m = \dfrac{2\pi}{\sqrt{5}}\, \text{s}\).