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

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.
Acceleration due to gravity $=g$
Centripetal acceleration $=v^2/R$
where,
v is the uniform speed of the car
R is the radius of the track
Effective acceleration $\left(g^{\prime }\right)$ is given as:

$g^{\prime }=\sqrt{g^2+\frac{v^4}{R^2}}$

$\therefore$ Time peroid, $t=2\pi \sqrt{\frac{l}{g^{\prime }}}$

$=2\pi \sqrt{\frac{l}{\sqrt{g^2+\frac{v^4}{R^2}}}}$