Question

A simple pendulum has a length $l$, mass of bob $m$. The bob is given a charge $q$. The pendulum is suspended between the vertical plates of the charged parallel plate capacitor. If $E$ is the field strength between the plates, then time period $T$ equal to:

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

Answer 1

The correct option is D $2\pi \sqrt{\frac{l}{\sqrt{g^2+{\left(\frac{qE}{m}\right)}^2}}}$

Here two force is applied - one is gravitational force $\left(mg\right)$ and other is electrostatic force on the pendulum bob. The resultant force , $\overrightarrow{F}=mg\hat{j}+qE\hat{i}$.
If $a_{eff}$ is the effective acceleration of pendulum bob , $m{\overrightarrow{a}}_{eff}=\overrightarrow{F}=mg\hat{j}+qE\hat{i}$.
$\therefore |{\overrightarrow{a}}_{eff}|=a_{eff}=\sqrt{g^2+(\frac{qE}{m}{)}^2}$
Now time period , $T=2\pi \sqrt{\frac{l}{a_{eff}}}=2\pi \sqrt{\frac{l}{\sqrt{g^2+(\frac{qE}{m}{)}^2}}}$