A simple pendulum of length l and a mass m of the bob is suspended in a car that is travelling with a constant speed v around a circle of radius R . If the pendulum undergoes small oscillations about its equilibrium position, the frequency of its oscillation will be

The centripetal acceleration on the bob as it oscillates (acting along the radius of
circle) $={\mathrm{V}}^2/\mathrm{R}.$ This will act horizontally towards the centre of circular path.
The total acceleration acting on the pendulum bob is therefore $a=\sqrt{{\mathrm{g}}^2+{\left(\frac{{\mathrm{v}}^2}{\mathrm{R}}\right)}^2}$ The frequency of oscillation will therefore be $\mathrm{n}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{a}}{\mathrm{\ell }}}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{{\left({\mathrm{g}}^2+\frac{{\mathrm{v}}^4}{{\mathrm{R}}^2}\right)}^{1/2}}{\mathrm{\ell }}}$