A particle is performing linear S.H.M. at a point A, on the path, its potential energy is three times kinetic energy. At another point B on the same path, its kinetic energy is 3 times the potential energy. The ratio of the potential energy at A to its potential energy at B is

Options

A. `9 : 1`
B. `1 : 9`
C. `1 : 3`
D. `3 : 1`

Answer

D

Explanation

`T.E._(A)=K.E_(A)+P.E_(A)`
`=KE_(A)+3KE_(A)`
`=4KE_(A)`
`T.E_(B)=KE_(B)+KE_(B)=KE_(B)+(1)/(3)KE_(B)`
`T.E_(B)=(4KE_(B))/(3)`