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Calculating frequency of oscillation

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A 4C point charge of mass 2kg is suspended by a string of length 6m. The charge is placed in the Earth's Gravitational field and a uniform horizontal electric field of strength $\sqrt{11}NC^{-2}$. If the charge is displaced slightly from equilibrium, what will be the frequency of oscillation? (g=10m/s)

I was trying to solve the question following way.

$$f=\frac{1}{2\pi} \sqrt{\frac{k}{m}}$$

$$f=\frac{1}{2\pi} \sqrt{\frac{F}{ml}}$$

$$f=\frac{1}{2\pi} \sqrt{\frac{Eq+mg}{ml}}$$

Where, $m$ is mass. $l$ is length. $E$ is electric field. $q$ is point charge. $g$ is gravitational acceleration. But, my answer was wrong. What am I missing?

The correct answer is $$\frac{\sqrt{2}}{2\pi}$$

But, I couldn't solve it anyway.

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2 comments

What's with the electric field strength in Newtons per square Coulomb? That's not the same as V/m as expected. Also, a uniform electric field will have a constant force on a point charge. Why do you expect a constant sideways force to have any effect on the pendulum frequency in the first place? Then there is g in units of speed, which also makes no sense. Olin Lathrop‭ 15 days ago

@Olin Lathrop Yes! I understood the question(maybe, 1 or 2 hours ago) after asking the question. Istiak‭ 15 days ago

1 answer

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While force is vector. So, we should use vector addition not "direct addition".

From vector addition formula

$$C=\sqrt{A^2+B^2+2ABcos\alpha}$$

For force,

$$F=Eq$$ $$F=mg$$ $$F=\sqrt{(Eq)^2+(mg)^2+2Eqmgcos\alpha}$$

$$f=\frac{1}{2\pi} \sqrt{\frac{\sqrt{(Eq)^2+(mg)^2+2Eqmgcos\alpha}}{ml}}$$

Then, problem solved.

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