Question:
Time period of a simple pendulum in a stationary lift is ‘T’. If the lift accelerates with \(\frac{g}{6}\) vertically upwards then the time period will be (Where g = acceleration due to gravity)
Time period of a simple pendulum in a stationary lift is ‘T’. If the lift accelerates with \(\frac{g}{6}\) vertically upwards then the time period will be (Where g = acceleration due to gravity)
Updated On: Mar 19, 2025
\(\sqrt{\frac{6}{5}}T\)
\(\sqrt{\frac{5}{6}}T\)
\(\sqrt{\frac{6}{7}}T\)
\(\sqrt{\frac{7}{6}}T\)
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The Correct Option is C
Solution and Explanation
\(T' = 2\pi \sqrt{\frac{I}{g_{\text{eff}}}}\)
\(T' = 2\pi \sqrt{\frac{I}{g + \frac{g}{6}}} = 2\pi \sqrt{\frac{6I}{7g}}\)
\(T' = \sqrt{\frac{6}{7}}T\)
So,the correct option is C
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Concepts Used:
Acceleration
In the real world, everything is always in motion. Objects move at a variable or a constant speed. When someone steps on the accelerator or applies brakes on a car, the speed of the car increases or decreases and the direction of the car changes. In physics, these changes in velocity or directional magnitude of a moving object are represented by acceleration.
