Question:
A simple pendulum is suspended in a car moving on a circular track of radius \( R \) with a uniform speed of \( \sqrt{1.732 g R} \). The pendulum is making small oscillations in a radial direction about its equilibrium position with a time period \( T \). If the car is at rest, the time period of the pendulum is:
A simple pendulum is suspended in a car moving on a circular track of radius \( R \) with a uniform speed of \( \sqrt{1.732 g R} \). The pendulum is making small oscillations in a radial direction about its equilibrium position with a time period \( T \). If the car is at rest, the time period of the pendulum is:
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For a pendulum in a moving car, the effective acceleration is the vector sum of the gravitational and centripetal accelerations, leading to a change in the time period of the pendulum.
Updated On: Mar 18, 2026
- \( T \)
- \( T / \sqrt{2} \)
- \( T \sqrt{3} \)
- \( 2T \)
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The Correct Option is B
Solution and Explanation
When the car is moving, the pendulum experiences both the gravitational and centripetal acceleration. The effective acceleration is:
\[
a_{\text{eff}} = \sqrt{g^2 + (1.732 g)^2} = 2g
\]
Thus, the time period of the pendulum when the car is moving is:
\[
T_{\text{moving}} = \frac{T}{\sqrt{2}}
\]
Hence, the time period of the pendulum when the car is moving is \( T / \sqrt{2} \).
Hence, the time period of the pendulum when the car is moving is \( T / \sqrt{2} \).
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