The ratio of kinetic energy of mean position to thhe potential energy when the displacement is half of the amplitude is :

Options

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

Answer

A

Explanation

Kinetic energy `E_(K)=(1)/(2)m omega^(2)(r^(2)-y^(2))`
At mean position `y=0,:. E_(K)=(1)/(2)m omega^(2)r^(2)`
Potential energy `E_(P)=(1)/(2)m omega^(2)y^(2)`
At ` y=( r )/(2),E_(P)=(1)/(2)m omega^(2)(r^(2))/(4)`.
`:." "(E_(K))/(E_(P))=(4)/(1)`
Correct choice is (a).