Question

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 ?

Answer 1

Correct Answer - In the given problem
`t = 2pisqrt((T)/(g_(e)))`
where `g_(e) ` is the effective value of acceleration due to gravity.
While the acceleration due to gravity `g` acts in the downward direction, the radial acceleration `(v^(2))/(R)` acts in the horizontal direction so, the effective value `g_(e)` of acceleration due to gravity is given by
`g_(e) = sqrt(g^(2) + ((v^(2))/(R))^(2))`
`:.T=2pisqrt((l)/(g^(2)+(v^(2)/(R))^(2)))` or `T=2pisqrt((l)/(g^(2)+(v^(2)/(R))^(2)))`