Q. A simple pendulum has a length $l$ and the mass of the bob is $m$. The bob is given a change $q$ coulomb. The pendulum is suspended between the vertical plates of a charged parallel plate capacitor. If $E$ is the electric field strength between the plates, the time period of the pendulum is given by
Explanation
(d) Time period of simple pendulum in air
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when it is suspended between vertical plates of a charged parallel plate capacitor, then acceleration due to electric field,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a = \frac{qE}{m}$
This acceleration is acting horizontally and acceleration due to gravity is acting vertically.
So, effective acceleration
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g^{''} = \sqrt{g^{2} + a^{2}} = \ \sqrt{g^{2} + \left( \frac{qE}{m} \right)^{2}}$
Hence, $T^{''} = \ 2\pi\frac{\sqrt{l}}{\sqrt{g^{2} + \left( \frac{qE}{m} \right)^{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.