The potential energy of a simple harmonic oscillator when the particle is half way to its end point is
(where, E is the total energy)

Options

A. `(1)/(4)E`
B. `(1)/(2)E`
C. `(2)/(3)E`
D. `(1)/(8)E`

Answer

A

Explanation

Potential energy of a simple harmonic oscillator
`U=(1)/(2)momega^(2)y^(2)`
Kinetic energy of a simple harmonic oscillator
`K=(1)/(2)momega^(2)(A^(2)-y^(2))`
Here, y=displacement from mean position
A=maximum displacement (or amplitude) from mean position
Total energy is
E=U+K
`=(1)/(2)momega^(2)y^(2)+(1)/(2)momega^(2)(A^(2)-y^(2))`
`=(1)/(2)momega^(2)A^(2)`
When the particle half way to its end point ie, at half of its amplitude then
`y=(A)/(2)`
Hence potential energy
`U=(1)/(2)momega^(2)((A)/(2))^(2)`
`=(1)/(4)((1)/(2)momega^(2)A^(2))`
`U=(E)/(4)`