Question

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 position, what will be its time period?

Answer 1

Hint: The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car. The acceleration acting on the bob will be the resultant of the two of the above accelerations.

Formula Used: The following formulas are used to solve this question.
 $\Rightarrow {a_{eff}} = \sqrt {{g^2} + {{\left( {\dfrac{{{v^2}}}{R}} \right)}^2}} $ where $ {a_{eff}} $ is the effective acceleration, $ g $ is the acceleration due to gravity, $ R $ is the radius of the circular track and $ v $ is the speed of the car.
 $\Rightarrow T = 2\pi \sqrt {\dfrac{l}{{{a_{eff}}}}} $ where $ T $ is the time period, $ l $ is the length of pendulum and $ {a_{eff}} $ is the effective acceleration.
 $\Rightarrow {a_{centripetal}} = \dfrac{{{v^2}}}{R} $ where $ {a_{centripetal}} $ is the centripetal acceleration.

Complete step by step answer
It is given that a simple pendulum of length $ l $ and having a bob of mass $ M $ is suspended in a car which is moving on a circular track.
The force necessary to keep an object moving in a curved path and that is directed inward toward the centre of rotation is the centripetal force.
Centripetal acceleration is the property of a body in motion, traversing a circular path. The acceleration is directed radially toward the centre of the circle and has a magnitude equal to the square of the body’s speed along the curve divided by the distance from the centre of the circle to the moving body. The force causing this acceleration is directed also toward the centre of the circle and is named centripetal force.
The circular track has a radius $ R $. The speed of the car is $ v $.
The car experiences centripetal acceleration $ {a_{centripetal}} $ where
 $\Rightarrow {a_{centripetal}} = \dfrac{{{v^2}}}{R} $.
The effective acceleration of a body $ {a_{eff}} $ is the resultant of the acceleration due to gravity $ g $ and centripetal acceleration $ {a_{centripetal}} $. The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car.
Thus, the effective acceleration is given by,
 $\Rightarrow {a_{eff}} = \sqrt {{g^2} + {{\left( {\dfrac{{{v^2}}}{R}} \right)}^2}} $
Time period of a pendulum is the time taken by a pendulum of length $ l $ to complete one oscillation.
It is known to us that, time period $ T = 2\pi \sqrt {\dfrac{l}{{{a_{eff}}}}} $ where $ l $ is the length of pendulum and $ {a_{eff}} $ is the effective acceleration.
Substituting the value of effective acceleration $ {a_{eff}} $ in the equation for time period $ T $.
 $\Rightarrow T = 2\pi \sqrt {\dfrac{l}{{{a_{eff}}}}} $
 $ \Rightarrow T = 2\pi \sqrt {\dfrac{l}{{\sqrt {{g^2} + {a_{centripetal}}^2} }}} $
Assigning, $ {a_{centripetal}} = \dfrac{{{v^2}}}{R} $ we get the time period as,
 $ \Rightarrow T = 2\pi \sqrt {\dfrac{l}{{\sqrt {{g^2} + {{\left( {\dfrac{{{v^2}}}{R}} \right)}^2}} }}} $
Thus, the time period of the pendulum is $ 2\pi \sqrt {\dfrac{l}{{\sqrt {{g^2} + {{\left( {\dfrac{{{v^2}}}{R}} \right)}^2}} }}} $.

Note
The centripetal acceleration is acting horizontally on the car. Acceleration due to gravity $ g $ is acting vertically downwards. Now the effective acceleration given by $ {a_{eff}} = \sqrt {{g^2} + {{\left( {\dfrac{{{v^2}}}{R}} \right)}^2}} $, is actually the effective acceleration due to gravity.