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Comments on Calculating frequency of oscillation

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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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General comments (2 comments)
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Olin Lathrop‭ wrote almost 3 years ago · edited almost 3 years ago

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.

deleted user wrote almost 3 years ago

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

Skipping 1 deleted comment.