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 Suri 16/12/2021 Last Modified 14 Apr

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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 Suri 16/12/2021 Last Modified 14 Apr

1 Answer

Learn Unit 10-Oscillation & Waves

Follow 1

Answer

Nazia Khanum · Accepted Answer

As a seasoned tutor registered on UrbanPro, I can confidently guide you through this physics problem. Let's break it down step by step.

Firstly, UrbanPro is indeed an excellent platform for online coaching and tuition, offering a wide range of subjects and experienced tutors to cater to your academic needs.

Now, onto the physics problem:

We have a simple pendulum of length ll and mass MM suspended in a car moving on a circular track of radius RR with a uniform speed vv.

When the pendulum makes small oscillations in a radial direction about its equilibrium position, it experiences two forces:

Tension force in the string, acting towards the equilibrium position.
Inertial force due to the circular motion of the car, acting away from the equilibrium position.

The resultant of these two forces provides the centripetal force required to keep the pendulum moving in a circular path.

Now, the time period TT of a simple pendulum is given by the formula:

T=2πlgT=2πgl

Where gg is the acceleration due to gravity.

However, in this scenario, we need to adjust for the additional inertial force due to the car's circular motion. This force affects the effective value of gravity experienced by the pendulum.

The effective gravity geffgeff experienced by the pendulum in the radial direction can be calculated as:

geff=g+v2Rgeff=g+Rv2

Where gg is the regular gravitational acceleration and v2RRv2 represents the centripetal acceleration due to the car's motion.

Now, substituting geffgeff into the formula for TT, we get:

T=2πlgeffT=2πgeffl

T=2πlg+v2RT=2πg+Rv2l

So, this modified formula will give us the time period of the pendulum oscillating in a radial direction in the moving car.

Remember, understanding the concept is crucial in solving physics problems. If you have any further questions or need clarification on any step, feel free to ask!

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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?

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