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The time period of a charged simple pendulum in an electric field is:

$\begin {array} {1 1} (a)\;\large\frac{1}{(2 \pi)} \sqrt{\large\frac{1}{mg}} & \quad (b)\;\large\frac{1}{(2\pi)} \sqrt{ \bigg[\large\frac{l}{(g-qE)\bigg]}} \\ (c)\;2 \pi \sqrt{\bigg[\large\frac{l}{\bigg(g – \large\frac{qE}{m} \bigg)}\bigg]} & \quad (d)\;2\pi\bigg[\large\frac{l}{(gm-qE)}\bigg] \end {array}$


1 Answer

Consider a simple pendulum of length $l$ suspended in a uniform electric field $E$.
Let the bob of dimple pendulum carry a charge $q$$
Net vertically downward force on the bob is $(mg-qE).$
Net acceleration $=\large\frac{( mg - qE)}{m}$$ = \large\frac{g - qE}{m}$
Time period, $2 \pi \sqrt{\bigg[\large\frac{l}{\bigg(g – \large\frac{qE}{m} \bigg)}\bigg]}$
Ans : (c)
answered Feb 28, 2014 by thanvigandhi_1

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