Gujarati
13.Oscillations
easy

There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is $T$. If the resultant acceleration becomes $g/4,$ then the new time period of the pendulum is

A

$0.8 \,T$

B

$0.25\, T$

C

$2\, T$

D

$4\, T$

Solution

(c) When lift is at rest, $T = 2\pi \sqrt {l/g} $

If acceleration becomes $g/4$ then

$T' = 2\pi \sqrt {\frac{l}{{g/4}}} = 2\pi \sqrt {\frac{{4l}}{g}} = 2 \times T$

Standard 11
Physics

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