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13.Oscillations
easy
A simple pendulum has time period 't'. Its time period in a lift which is moving upwards with acceleration $3 ms ^{-2}$ is
A
$t \sqrt{\frac{9.8}{12.8}}$
B
$t \sqrt{\frac{12.8}{9.8}}$
C
$t \sqrt{\frac{9.8}{6.8}}$
D
$t \sqrt{\frac{6.8}{9.8}}$
Solution
(a)
$t \propto \frac{1}{\sqrt{9.8}}, t ^{\prime} \propto \frac{1}{\sqrt{12.8}}$
$\left(\because g ^{\prime}=9.8+3=12.8\right)$
$\therefore \frac{ t ^{\prime}}{ t }=\sqrt{\frac{9.8}{12.8}} \Rightarrow t ^{\prime}=\sqrt{\frac{9.8}{12.8}} t$
Standard 11
Physics
