Q. A simple pendulum has a length $l$ and the mass of the bob is $m$. The bob is given a charge $q$ 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
A) $2\pi\sqrt{\frac{l}{g}}$
B) $2\pi\sqrt{\frac{l}{\sqrt{g + \frac{qE}{m}}}}$
C) $2\pi\sqrt{\frac{l}{\sqrt{g - \frac{qE}{m}}}}$
D) $2\pi\sqrt{\frac{l}{\sqrt{g^{2} + \left( \frac{qE}{m} \right)^{2}}}}$
Explanation
(d) Time period of simple pendulum in air
$T = 2\pi\sqrt{\frac{l}{g}}$
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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, $T^{''} = 2\pi\sqrt{\frac{l}{\sqrt{g^{2} + \left( \frac{qE}{m} \right)^{2}}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.