A simple pendulum of length 'I' 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 oscillation in a radial direction about its equilibrium position, then what will be its time period?

Explanation

Let `theta` be the angle of banking
period of oscillation `T=T=2pisqrt((Lcos theta)/(g))`
`"but "tan theta=(v^(2))/(Rg)`, so that
`AC=sqrt(v^(4)+R^(2)g^(2))`
`"and "cos theta=(BC)/(AC)=(Rg)/(sqrt(v^(4)+R^(2)g^(2)))`

`"or "cos theta=(g)/(sqrt((v^(4))/(R^(2))+g^(2)))`
Hence the period of oscillation of pendulum will be,
`T=2pi sqrt((Lg)/(g sqrt((v^(4))/(R^(2))+g^(2))))" or "T=2pisqrt((L)/(sqrt((v^(4))/(R^(2))+g^(2))))`