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

Answer 1

Answer: (D)

Time period of simple pendulum in air
$T=2\pi \sqrt{\frac{l}{g}}$ 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 $g^{\prime }=\sqrt{g^2+a^2}=\sqrt{g^2+{\left(\frac{qE}{m}\right)}^2}$ Hence, $T^{\prime }=2\pi$ $\sqrt{\frac{l}{\sqrt{g^2+{\left(\frac{qE}{m}\right)}^2}}}$