Question

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 capacitor. If $E$ is electric field strength between the plates, the time period of pendulum is given by

• A. $2\pi \sqrt{\frac{1}{\sqrt{g-\frac{qE}{m}}}}$
• B. $2\pi \sqrt{\frac{l}{\sqrt{g^2+{\left(\frac{qE}{m}\right)}^2}}}$
• C. $2\pi \sqrt{\frac{l}{g}}$
• D. $2\pi \sqrt{\frac{1}{\sqrt{g+\frac{qE}{m}}}}$

Answer 1

Answer: (B)

Time period of simple pendulum in air
$T=2\pi \frac{\sqrt{l}}{g}$

When it is suspended between vertical plates of a charged parallel plate capacitor, then accelertion due to electric field,
$a=\frac{qE}{m}$
This acceleration is acting horizontally and acceleration due to gravity is acting vertically. So, effective acceleration
$g^{\prime }=\sqrt{g^2+a^2}=\sqrt{g^2+{\left(\frac{qE}{m}\right)}^2}$
Hence, $T^{\prime }=2\pi \sqrt{\frac{l}{\sqrt{g^2+{\left(\frac{qE}{m}\right)}^2}}}$