A simple pendulum suspended from the roof off a lift oscillates with frequency `upsilon` when the lift is at rest. If the lift falls freely under gravity, its frequency of oscillation becomes

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

D

Explanation

At mean position,
`mg+(mv^(2))/(r)=3mgrArrv=sqrt(2gl)`…..(i)
and if the body displace by angle `theta` with vertical, then
`v=sqrt(2gl(1-costheta))`……(ii)
On comparing Eqs. (i) and (ii), we get
`costheta=0rArrtheta=90^(@)`