A simple pendulum of length l and having a bob of mass M is suspended ina car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium, what will be its time period ?

Explanation

Time period of a pendulum is given by ,
`T = 2pi sqrt(("displacement")/("acceleration"))`
In this case, the pendulum has two acceleration:
(i) acceleration due to gravity .g.
(ii) centripetal acceleration `a_(c ) = (v^(2))/(R )`
Since g and `(v^(2))/(R )` are at right angle to each other, net acceleration,
`a = sqrt(g^(2) + ((v^(2))/(R ))^(2)) = (g^(2) + (v^(4))/(R^(2)))^(1//2)`