Question

The bob of a simple pendulum is released when its string is horizontal. If kinetic energy of bob is $0.1732J$ when string makes an angle $30°$ with horizontal, then its kinetic energy when string makes an angle $60°$ with horizontal is :

• A. $0.2J$
• B. $0.3J$
• C. $0.4J$
• D. $0.5J$

Answer 1

Answer: (B)

$0.3J$

Let the length of string be $lm$

When string makes ${30}^{\circ }$ with horizontal then the vertical drop of pendulum is $lsin\left({30}^{\circ }\right)=l/2m$

$K.E.gain=P.E.loss$

$\Rightarrow mg\frac{l}{2}=0.1732J=>mgl=0.1732\times 2J$

When string makes ${60}^{\circ }$ with horizontal then the vertical drop of pendulum is $lsin\left({60}^{\circ }\right)=\frac{l\sqrt{3}}{2}m$

K.E Gain = P.E. loss = $mg\frac{l\sqrt{3}}{2}=0.1732\times 2\times \frac{\sqrt{3}}{2}$ = $0.3J$