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 $
BETA
