The ratio 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. `4:1`
B. `8:1`
C. `3:1`
D. `2:1`

Answer

C

Explanation

The kinetic energy of the particle executting SHM at a distance x from its equilibrium position is
`K=(1)/(2)momega^(2)(A^(2)-x^(2))`
and the potential energy is
`U=(1)/(2)momega^(2)x^(2)`
where A is the amplitude, `omega` is the angular frequency and M is the mass of the body .
At `x=(A)/(2)`
`K=(1)/(2)momega^(2)(A^(2)-((A)/(3))^(2))`
and `U=(1)/(2)momega^(2)((A)/(2))^(2)`
Their corresponding ratio is
`(K)/(U)=((1)/(2)momega^(2)((A^(2)-(A)/(2))^(2)))/((1)/(2)momega^(2)((A)/(2))^(2))=((A^(2)-(A^(2))/(4)))/((A^(2))/(4)`
`=((3)/(4)A^(2))/((A^(2))/(4))=(3)/(1)`