Question

A simple pendulum has a length $l$, mass of bob $m$. The bob is given a charge $q$ coulomb. The pendulum is suspended in a uniform horizontal electric field of strength $E$ as shown in figure, then calculate the time period of oscillation when the bob is slightly displaced from its mean position:

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

Now two forces are acting on the bob.
Force due to gravity $=mg$
Force due to electric field $=qE$
Both the forces are perpendicular to each other.
So net force on the bob, $F_{net}=\sqrt{m^2{g}^2+q^2{E}^2}$

So, effective acceleration, $g_{eff}=\frac{F_{net}}{m}=\sqrt{g^2+{\left(\frac{qE}{m}\right)}^2}$

Time period of simple pendulum, $T=2\pi \sqrt{l/g_{eff}}$

Therefore, time period wil be
$T=2\pi \sqrt{\frac{l}{\sqrt{g^2+{\left(\frac{qE}{m}\right)}^2}}}$