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 $v_1$ and $v_2$ 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. $\sqrt{\frac{2gx(1-\cos {\theta }_1)}{1-\cos {\theta }_2}}$
• D. $\sqrt{\frac{1-\cos {\theta }_1}{2gx(1-\cos {\theta }_2)}}$

Answer 1

The correct option is B $\sqrt{\frac{1-\cos {\theta }_1}{1-\cos {\theta }_2}}$

By energy conservation,

$v=\sqrt{2g(x-x\cos \theta )}$

Similarly, $v_1=\sqrt{2gx(1-\cos {\theta }_1)}$

$v_2=\sqrt{2gx(1-\cos {\theta }_2)}$

$\frac{v_1}{v_2}=\sqrt{\frac{1-\cos {\theta }_1}{1-\cos {\theta }_2}}$