Calculating frequency of oscillation
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.
1 answer
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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