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?

Asked by Abhisek | 1 year ago |  85

##### Solution :-

The bob of the simple pendulum will experience the centripetal acceleration provided by the circular motion of the car and the acceleration due to gravity.

Acceleration due to gravity = g

Centripetal acceleration = $$\dfrac{v^2}{4}$$

Where,

v is the uniform speed of the car

R is the radius of the track

Effective acceleration (g) is given as

g = $$\sqrt{g^2}+\dfrac{V^4}{R^2}$$

Hence,

Time period, t = $$2π\dfrac{\sqrt{I}}{g'}$$

$$2π \sqrt{\dfrac{I}{g^2}+\dfrac{v^4}{R^2}}$$

Therefore, its time period will be $$2π \sqrt{\dfrac{I}{g^2}+\dfrac{v^4}{R^2}}$$

Answered by Pragya Singh | 1 year ago

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