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

1 Answer

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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