A particle performing S.H.M. about their mean position with the equation of velocity is given by `4v^(2)=25-x^(2)`, then the period of motion is

Options

A. `2pi`
B. `pi`
C. `3pi`
D. `4pi`

Answer

D

Explanation

`4v^(2)=25-x^(2)`
`v^(2)=(1)/(4)[25-x^(2)]`
`v=pm(1)/(2)sqrt(25-x^(2))`
Compare it with standard equation of velocity,
`v=omegasqrt(A^(2)-x^(2))`
`omega=(1)/(2) therefore T=(2pi)/(omega)=(pi)/(1//2)`
`therefore T=4pis`