Question

A simple pendulum of length 1 m has a bob of mass 200 g. It is displaced ${60}^{\circ }$ and then released. Find the kinetic energy of the bob when
It passes through the mean position

• A. $1J,1.732J$
• B. $2J,1.732J$
• C. $2J,0.732J$
• D. $1J,0.732J$

Answer 1

The correct option is D $1J,0.732J$

Let the length of simple pendulum be $=1\text{m}$

$\text{Mass}=2wgm=0.2kg$

It is moved by an angle of 60 for vertical height of pendulum at the beginning

= length $(2-ws60)$

$=1(1-0.5)$

$=0.5\text{m}$

$\text{Potential energy}=\text{mgh}$

$=0.2\times 10\times 0.5=1J$

when it is released and it reaches mean position, the potential energy at beginning gets changed to kinetic energy.

Therefore, Kinetic Energy of bob at mean point $=1J$.

Therefore, the correct option is (a)$1J$.

To find: kinetic energy at position ${30}^{\circ }$

Explanation: $h=l\cos 30-l\cos 60$

$=\frac{l\sqrt{3}}{2}-l/2=\frac{l(\sqrt{3}-1)}{2}$

By applying the conservation of mechanical energy.

Gain in kinetic energy= drop in potential energy.

$K{E}_{{30}^{\circ }}=mgh=.2kg\times 10\times \frac{(\sqrt{3}-1)}{2}$

$=0.7321$