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Home » A simple pendulum of length I and having a bob of mass is suspended in a car. The car is moving on a circular track of radius with a uniform speed If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?
A simple pendulum of length I and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius \mathbf{R} with a uniform speed \mathbf{v} . If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?
CBSE | Class 11 | NCERT | Oscillations | Physics
A simple pendulum of length I and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius \mathbf{R} with a uniform speed \mathbf{v} . If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

The centripetal acceleration supplied by the circular motion of the car, as well as the acceleration due to gravity, will be felt by the bob of the basic pendulum.

Acceleration due to gravity is nknown as \mathrm{g}

Centripetal acceleration is known as v^{2} / R

Where,

v= uniform speed of the car

\mathrm{R}= radius of the track

Effective acceleration \left(\mathrm{g}^{\prime}\right) is given as

g^{\prime}=\sqrt{g}^{2}+\left(v^{4} / R^{2}\right)

So,

Time period, t=2 \pi \sqrt{l} / \mathrm{g}

=2 \pi \sqrt{\left(l / \sqrt{g}^{2}+v^{4} / R^{2}\right)}

As a result, the time period will be \left.2 \pi \sqrt{(} \mathrm{l} / \mathrm{Vg}^{2}+\mathrm{v}^{4} / \mathrm{R}^{2}\right)

i

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