If displacement `x` and velocity `v` related as
`4v^(2) = 25 - x^(2)m` in a `SHM` Then time period of given `SHM` is (consider SI unit)

Options

A. `6pi`
B. `4pi`
C. `3pi`
D. `2pi`

Answer

B

Explanation

`4upsilon^(2)=25-x^(2)`
Differentiating it w.r.t. to t, we have
`4xx2upsilon(dupsilon)/(dt)=-2x(dx)/(dt)`
or `4xx(dupsilon)/(dt)=-x` `( :' (dx)/(dt)=upsilon)`
or `4a=-x or a=-(x)/(4)` …(i)
Comparing (i) with `a=-omega^(2)x, ` we have
`omega^(2)=(1)/(4), T=(2pi)/(omega)=(2pi)/((1//2))=4pi`