There is a simple pendulum hanging from ceiling of a lift. When lift is stand still, time period of

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13.Oscillations

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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) $T \propto \frac{1}{{\sqrt g }}$

==> $\frac{{{T_2}}}{{{T_1}}} = \sqrt {\frac{{{g_1}}}{{{g_2}}}} $

$ = \sqrt {\left( {\frac{g}{{g/4}}} \right)} $

==> ${T_2} = 2{T_1} = 2T$

Standard 11

Physics

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