A simple pendulum of length l and having a bob of mass M is suspended in a 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 ?

Options

A. `T=2pisqrt((L)/(g))`
B. `T=2pisqrt((L)/(sqrt(g^(2)+(v^(4))/(R^(2)))))`
C. `T=2pisqrt((L)/(sqrt(g^(2)+(v^(2))/(R))))`
D. `T=2pisqrt((L)/(g^(2)-(v^(4))/(R^(2))))`

Answer

B

Explanation

The bob of the pendulum has two accelerations:
(i) centripetal acceleration, `a_(c)=(v^(2))/(R)`
it acts horizontally.
(ii) Acceleration due to gravity=g. it acts vertically downwards.
The effective acceleration due to gravity.
`g'=sqrt(g^(2)+a_(c)^(2))=sqrt(g^(2)+((v^(2))/(R))^(2))=sqrt(g^(2)+(v^(4))/(R^(2)))`
`therefore`time period, `T=2pisqrt((L)/(g'))=2pisqrt((L)/(sqrt(g^(2)+(v^(4))/(R^(2)))))`