A particle starts oscillating simple harmonically from its equilibrium position then, the ratio of kinetic energy and potential energy of the particle at the time `T//12` is: `(T="time period")`

Options

A. `1 : 2`
B. `2 : 1`
C. `1 : 3`
D. `3 : 1`

Answer

D

Explanation

The particle starts from the mean position
`therefore x= A "sin" omega t=A "sin" (2pi t)/(T)`
`therefore " When " t=(T)/(12)`
`x=A "sin" (2pi )/(T)xx(T)/(12)=A "sin" (pi)/(6)=(A)/(2)`
`therefore K.E. =(1)/(2)mv^(2)=(1)/(2)m omega^(2)(A^(2)-x^(2))`
`=(1)/(2)m omega^(2)(A^(2)-(A^(2))/(4))`
`=(3)/(4)((1)/(2)m omega^(2)A^(2))`
and `P.E. =(1)/(2)m omega^(2)x^(2)=(1)/(4)((1)/(2)m omega^(2)A^(2))`
`because x^(2)=(A^(2))/(4)`
`therefore (K.E.)/(P.E.)=((3)/(4)((1)/(2)m omega^(2)A^(2)))/((1)/(4)((1)/(2)momega^(2)A^(2)))=(3)/(1)`