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?
T = 2π)
Explanation
The car moving on a circular track with uniform speed v and radius R has centripetal acceleration v²/R directed towards the center. In the non-inertial frame of the car, the pendulum bob experiences two accelerations: gravitational acceleration g (vertically downward) and centrifugal acceleration v²/R (radially outward). The effective acceleration g_eff is the resultant of these two perpendicular accelerations. Using the formula T = 2π from the context and substituting g_eff for g gives the time period for small oscillations about the equilibrium position.
Solution Steps
Step 1: In the car's non-inertial frame, the pendulum bob experiences gravitational acceleration g (vertically downward) and centrifugal acceleration v²/R (radially outward).
Step 2: The effective acceleration is the resultant of these two perpendicular accelerations: g_eff = ²) = .
Step 3: Using the formula T = 2π from the context and replacing g with g_eff: T = 2π = 2π).