Question

The time period of small oscillations of a uniform rod of length L pivoted about one end and oscillating in a vertical plane is $T_1$ and time period of a simple pendulum of same length L is $T_2$ . The ratio $\frac{T_1}{T_2}$ is

• A. $\frac{2}{\sqrt{3}}$
• B. $\sqrt{\frac{1}{3}}$
• C. $\sqrt{\frac{2}{3}}$
  
• D. $\sqrt{\frac{3}{4}}$

Answer 1

The correct option is D $\sqrt{\frac{3}{4}}$

$x=x_{0}si{n}^2\omega t=\frac{x_0}{2}(1-cos2\omega t)=\frac{x_0}{2}-\frac{x_0}{2}cos2\omega t$
Frequency ${\omega }^{\prime }=2\omega$
$\Rightarrow \frac{2\pi }{T^{\prime }}=2\omega \Rightarrow T^{\prime }=\left(\frac{\pi }{\omega }\right)Amplitude=\frac{x_0}{2}$