When the car moves along a circular track, the bob of the simple pendulum will possess centripetal acceleration, $a_c = \frac{v^2}{R}$ (radially inward along horizontal) The acceleration due to gravity (g) acts vertically downwards. Therefore, effective acceleration of the pendulum, $g' = \sqrt{g^2 + \left(\frac{v^2}{R}\right)^2}$$= \sqrt{g^2 + \frac{v^4}{R^2}}$Hence, time period of the simple pendulum,$T = 2\pi \sqrt{\frac{l}{g'}}$$= 2\pi \sqrt{\frac{l}{g^2 + \frac{v^4}{R^2}}}$