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13.Oscillations
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The period of a simple pendulum measured inside a stationary lift is found to be $T$. If the lift starts accelerating upwards with acceleration of $g/3,$ then the time period of the pendulum is
A
$\frac{T}{{\sqrt 3 }}$
B
$\frac{T}{3}$
C
$\frac{{\sqrt 3 }}{2}T$
D
$\sqrt 3 \,T$
Solution
(c) For stationary lift ${T_1} = 2\pi \sqrt {\frac{l}{g}} $
For ascending lift with acceleration a, ${T_2} = 2\pi \sqrt {\frac{l}{{g + a}}} $
$ \Rightarrow \frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{g + a}}{g}} $
$\Rightarrow \frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{g + \frac{g}{3}}}{g}} = \sqrt {\frac{4}{3}}$
$\Rightarrow {T_2} = \frac{{\sqrt 3 }}{2}T$
Standard 11
Physics
