Question

A simple pendulum is suspended in a lift which is going up with an acceleration of $5{\text{m/s}}^2$. An electric field of magnitude $5N/C$ and directed vertically upward is also present in the lift. The charge of the bob is $1\mu C$ and mass is $1mg$. Taking $g={\pi }^2$ and length of the simple pendulum $1m$, find the time period of the simple pendulum (in sec).

Answer 1

We know that time period of a simple pendulum is given by
$T=2\pi \sqrt{\frac{l}{g}}$
But in this case $g$ is not the same. So we have to take ${}^{\prime }{g}^{\prime }$ effective ($g_{eff}$)
$T=2\pi \sqrt{\frac{l}{g_{eff}}}$
Given that acceleration of the lift is $5{\text{m/s}}^2$
As acceleration of the lift and acceleration due to electric field is in same direction. So we consider relative acceleration as
$=\frac{qE}{M}-5$

$g_{eff}=g-(\frac{qE}{M}-5)=g-\frac{qE}{M}+5g_{eff}={\pi }^2+5-\frac{1\times 5\times {10}^{-6}}{1\times {10}^{-6}}$
$g_{eff}={\pi }^2$

Now,
$T=2\pi \sqrt{\frac{l}{{\pi }^2}}=2\text{sec}$