Question

A small metallic bob of mass m of a simple pendulum of length L is suspended between the two oppositely charged large parallel plates. The electric field E act vertically. If positive charge 'q' is given to bob and displaced slightly from mean position then time period of oscillation will be

Answer 1

Answer:$T = 2\pi \sqrt\dfrac{l}{\sqrt{(g)^{2} + (\dfrac{qE}{m})^{2}}}$

Explanation:

Given that,

Mass = m

Length = l

Positive charge = q

We know that,

Two forces acting on the bob.

First electric force in horizontal due to electric field

$F_{1} = qE$

Second force in vertically

$F_{2} = mg$

Now, net force is

$F = \sqrt{(F_{1})^{2} +(F_{2})^{2}}$

$F = \sqrt{(qE)^{2} +(mg)^{2}}$

Now, according to newton's second law

F = mg

$g = \dfrac{F}{m}$

Where,

F = net force

g = acceleration due to gravity

Now, the effective value of acceleration due to gravity

$g _{net} = \sqrt \dfrac{(F_{net})^{2}}{(m)^{2}}$

$g_{net} = \sqrt \dfrac{(mg)^{2}+ (qE)^{2}}{m^{2}}$

$g_{net} = \sqrt{(g)^{2} + (\dfrac{qE}{m})^{2}}$

now, the time period is

$T = 2\pi \sqrt\dfrac{l}{g_{net}}$

$T = 2\pi \sqrt\dfrac{l}{\sqrt{(g)^{2} + (\dfrac{qE}{m})^{2}}}$

Hence, the time period of oscillation will be $T = 2\pi \sqrt\dfrac{l}{\sqrt{(g)^{2} + (\dfrac{qE}{m})^{2}}}$