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Q.
The period of oscillation of a simple pendulum is $T$ in a stationary lift. If the lift moves upwards with acceleration of $8 \,g$, the period will:
AFMCAFMC 2004
Solution:
When lift moves upwards net acceleration of lift increases.
When lift moves upwards that net force on it is in the upward direction.
Therefore, from Newton's second law, the net force
$F - mg = ma$
$\Rightarrow F = m(g + a)$
Given, $a=8 g$
$\therefore g' = g + 8 g = 9 g$
Also time period $(T)=2 \pi \sqrt{\frac{l}{g}}$
$\Rightarrow T \propto \frac{1}{\sqrt{g}}$
Since $T^{2} g=$ constant
$\therefore T_{1}^{2} g=T_{2}^{2} \times 9 g$
$\Rightarrow T_{2}=\frac{T_{1}}{3}=\frac{T}{3}$
