The ration of kinetic energy to the potential energy of a particle executing SHM at a distance equal to half its amplitude , the distance being measured from its equilibrium position is

Options

A. `2 : 1`
B. `3 : 1`
C. `8 : 1`
D. `4 : 71`

Answer

B

Explanation

Potential energy of a particle executing simple harmonic motion also periodic with period T/2. So, potential energy is zero at the mean position and maximum at the extreme displacements.
Let the amplitude of SHM be A.
Now, potential energy of SHM `= (1)/(2)kx^(2)`. Here, `x = (A)/(2)`
`U=(1)/(2)k(A^(2))/(4)" "...(i)`
Kinetic energy, `K=(1)/(2)kA^(2)-(1)/(2)k(A^(2))/(4)=(3)/(8)kA^(2)" "...(ii)`
On dividing Eq. (ii) by Eq. (i), we get
`(K)/(U)=(3)/(8)xx(8)/(1)=(3)/(1)`