$A$ 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?

(N/A) The bob of the simple pendulum experiences two accelerations: the acceleration due to gravity $(g)$ acting vertically downwards and the centripetal acceleration $(a_c = v^2/R)$ acting horizontally towards the center of the circular track.
The effective acceleration $(a_{\text{eff}})$ is the vector sum of these two perpendicular accelerations:
$a_{\text{eff}} = \sqrt{g^2 + a_c^2} = \sqrt{g^2 + (v^2/R)^2}$
The time period $(T)$ of a simple pendulum is given by the formula:
$T = 2\pi \sqrt{\frac{l}{a_{\text{eff}}}}$
Substituting the value of $a_{\text{eff}}$:
$T = 2\pi \sqrt{\frac{l}{\sqrt{g^2 + (v^2/R)^2}}}$