Home
Class 12
PHYSICS
The total energy of the particle executi...

The total energy of the particle executing simple harmonic motion of amplitude A is 100 J. At a distance of 0.707 A from the mean position, its kinetic energy is

A

25 J

B

50 J

C

100 J

D

`12.5 J`

Text Solution

Verified by Experts

The correct Answer is:
B
We know that,
Total energy
`E = 1/2 momega ^(2) A ^(2) = 100 J`
and kinetic energy KE
`=1/2 m omega ^(2) (A ^(2) - Y ^(2))`
`= 1/2 momega ^(2) [ A ^(2) - ((A)/( sqrt2)) ^(2) ]`
`= 1/2 m omega ^(2) [ ( 2 A ^(2) - A ^(2))/( 2) ]`
`=1/2 m omega ^(2) (A ^(2))/( 2)`
`= (100)/(2) J = 50 J`
Promotional Banner

Similar Questions

Explore conceptually related problems

A particle executes SHM of amplitude A .At what distance from the mean position is its kinetic energy equal to its potential energy?

The motion that repeat themselves are called periodic motions but for a simple harmonic motion, the force must be proportional to the displacement and it is directed towards the centre of motion. A particle executes a SHM of amplitude 'a'. At what distance from the mean position is its kinetic energy equal to its potential energy?