Question

A simple pendulum of length \(L\) is placed between the plates of a parallel plate capacitor having electric field \(E\), as shown in figure. Its bob has mass \(m\) and charge \(q\). The time period of the pendulum is given by :


bbb bbb

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

Answer 1

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

Given:
Pendulum length $=l$
Electric field $=E$
Bob mass $=m$
Charge on Bob $=q$

Time period of the pendulum $\left(T\right)$ is given by
$T=2\pi \sqrt{\frac{L}{g_{eff}}}$

$g_{eff}=\frac{\sqrt{\left(m{g}^2\right)+\left(q{E}^2\right)}}{m}$

$\Rightarrow g_{eff}=\sqrt{g^2+{\left(\frac{gE}{m}\right)}^2}$

$\therefore T=2\pi \sqrt{\frac{L}{\sqrt{g^2+{\left(\frac{qE}{m}\right)}^2}}}$.

Hence option $\left(D\right)$ is correct.