Question

A simple pendulum with bob of mass $m$ and length $x$ is held in position at an angle $\theta_{1}$ and then angle $\theta_{2}$ with the vertical. When released from these positions, speeds with which it passes the lowest positions are $v_{1}$ & $v_{2}$ respectively. Then, $\frac{v_{1}}{v_{2}}$ is

• A. $\frac{1-\cos \theta_{1}}{1-\cos \theta_{2}}$
• B. $\sqrt{\frac{1-\cos \theta_{1}}{1-\cos \theta_{2}}}$
• C. $\sqrt{\frac{2 g x\left(1-\cos \theta_{1}\right)}{1-\cos \theta_{2}}}$
• D. $\sqrt{\frac{1-\cos \theta_{1}}{2 g x\left(1-\cos \theta_{2}\right)}}$

Answer 1

Answer: (B)

$U_{i}+k_{i}=U_{f}+k_{f}$ (Mechanical energy conservation)
$m g l(1-\cos \theta)=\frac{1}{2} m v^{2}$
$\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{1-\cos \theta_{1}}{1-\cos \theta_{2}}$
$\frac{v_{1}}{v_{2}}=\sqrt{\frac{1-\cos \theta_{1}}{1-\cos \theta_{2}}}$