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?

A simple pendulum of length 'I' and having a bob of mass M is suspende

1.	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?
Answer» Solution :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))))`