Laws Of Motion Question 207

Question: The time period of a simple pendulum measured inside a stationary lift is found to be T. If the lift starts accelerating upwards with an acceleration g/3, the time period is [EAMCET 1994; CMEET Bihar 1995; RPMT 2000]

Options:

A) $ T\sqrt{3} $

B) $ T\sqrt{3}/2 $

C) $ T/\sqrt{3} $

D) $ T/3 $

Show Answer

Answer:

Correct Answer: B

Solution:

$ T=2\pi \sqrt{\frac{l}{g}} $ and $ T’=2\pi \sqrt{\frac{l}{4g/3}} $

$ [As\ g’=g+a=g+\frac{g}{3}=\frac{4g}{3} $ ] $ T’= $

$ \frac{\sqrt{3}}{2}T $



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