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) $2\pi\sqrt{\frac{l}{\sqrt{g^{2}+[\frac{qE}{m}]^{2}}}}$

Answer 1

answer : option (d) $2\pi\sqrt{\frac{l}{\sqrt{g^{2}+[\frac{qE}{m}]^{2}}}}$

explanation : you should remember that time period of simple pendulum is given by, $T=2\pi\sqrt{\frac{l}{a_{eff}}}$

where, $\bf{a_{eff}}$ means effective acceleration or net acceleration acting on bob and l is length of simple pendulum.

due to presence electric field, acceleration of bob , $a_E=\frac{qE}{m}$ along positive direction of x - axis.

and acceleration of bob due to gravity, $a_g=g$ along vertically downward direction.

so, $a_{eff}=\frac{qE}{m}\hat{i}-g\hat{j}$

hence, magnitude of $a_{eff}=\sqrt{\left(\frac{qE}{m}\right)^2+g^2}$

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