A simple pendulum has a length l and the mass of the bob is m. The bob is given a charge q coulomb. The pendulum is suspended between the vertical plate of a charged parallel plate capacitor. If E is the electric field strength between the plates, the time period of the pendulum is given by

Options

A. `2pi sqrt((l)/(g))`
B. `2pi sqrt((l)/(sqrt(g+(qE)/(m))))`
C. `2pi sqrt((l)/(sqrt(g-(qE)/(m))))`
D. `pi sqrt((l)/(sqrt(g^(2)+((qE)/(m))^(2))))`

Answer

D

Explanation

Time period of a simple pendulum,
`T = 2pi sqrt((l)/(g))`,
Acceleration due to electric field,
`a=(qE)/(m)`
Acceleration is acting horizontally.
Acceleration due to gravity is acting vertically.
Therefore effective acceleration,
`g'=sqrt(g^(2)+a^(2))=sqrt(g^(2)+((qE)/(m))^(2))`
`therefore T=2pi sqrt((l)/(sqrt(g^(2)+((qE)/(m))^(2))))`