If the displacement (x) and velocity (v) of a particle executing simple harmonic motion are related through the expression `4v^2=25-x^2`, then its time period is given by

Options

A. `pi`
B. `2pi`
C. `4pi`
D. `6pi`

Answer

B

Explanation

`4v^(2)=25-x^(2)`
Differentiating it both sides, we get
`4(2v(dv)/(dt))=-2x(dx)/(dt)or4(dv)/(dt)=-x (because (dx)/(dt)=v)`
or `4a=-xora=-(1)/(4)x(because(dv)/(dt)=a)`
Comparing it with `a=-omega^(2)x`, we get `omega^(2)=(1)/(4)oromega=(1)/(2)`
`therefore` Time period, `T=(2pi)/(omega)=(2pi)/(1//2)=4pi`