A simple pendulum of length `l` is suspended in a car that is travelling with a constant speed `v` around a circle of radius `r`. If the pendulum undergoes small oscillations about its equilibrium position , what will its freqeuency of oscillation be ?

Options

A. `(1)/(2pi)sqrt((g)/(1))`
B. `(1)/(2pi)sqrt((g)/(R))`
C. `(1)/(2pi)sqrt((g^(2)+(v^(4))/(R^(2)))^(1//2)/(1)`
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’s bob is therefore
`a=sqrt(g^(2)+((v^(2))/(R))^(2))`
The frequency of oscillation will therefore be
`n=(1)/(2pi)sqrt((a)/(1))=(1)/(2pi)sqrt(((g^(2)+(v^(4))/(R^(2))^(1//2))/(1))`