A simple pendulum of length `l` and mass `m` is suspended in a car that is moving with constant speed `v` around a circle of radius `r`. Find the period of oscillation and equilibrium position of the pendulum.

Options

A. `(1)/(2pi)sqrt((g)/(l))`
B. `(1)/(2pi)sqrt((g)/(R ))`
C. `(1)/(2pi)sqrt(((g^2+(v^4)/(R^2)))/(l))`
D. `(1)/(2pi)sqrt((v^2)/(Rl))`

Answer

C

Explanation

The centripetal acceleration on the bob as it oscillates (acting along the radius of circle)`=(v^2)/(R )`. This will act horizontally towards the centre of circular path.
The total acceleration acting on the pendulum bob is therefore `a=sqrt(g^2+((v^2)/(R ))^2)`
The frequency of oscillation will therefore be
`n=(1)/(2pi)sqrt((a)/(l))=(1)/(2pi)sqrt(((g^2+(v^4)/(R^2))^((1)/(2)))/(l))`