Question

The bob of a simple pendulum has a mass of 40 g and a positive charge of $4.0\times {10}^{-6}C.$ It makes 20 oscillations in 45 s. A vertical electric field pointing upward and of magnitude $2.5\times {10}^{4}N{C}^{-1}$ is switched on. How much time will it now take to complete 20 oscillations?

Answer 1

$T_1=2\pi \sqrt{\frac{l}{g}}a=\frac{qE}{m}=\frac{4\times {10}^{-6}\times 2.5\times {10}^4}{40\times {10}^{-3}}=2.5m/s^2\text{When no electric field is applied}{T}_2=2\pi \sqrt{\frac{1}{g-a}}$
Where $a=\frac{qE}{m}$ upward accelertion of mass due to electric field.
$\frac{T_2}{T_1}=\sqrt{\frac{g}{g-a}}+\sqrt{\frac{g}{g-\frac{qE}{m}}}=\sqrt{\frac{9.8}{9.8-0.5}}{T}_2=45\times \sqrt{\frac{9.8}{7.3}}=52$