Question

A particle of mass $m$ and charge $ - q$ performs SHM in a tunnel along the diameter of a uniformly charged sphere of radius $R$ with the total charge $Q$. The angular frequency of the particle’s simple harmonic motion, if its amplitude $ < R$, is given by
A. $\sqrt {\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{qQ}}{{mR}}} $
B. $\sqrt {\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{qQ}}{{m{R^2}}}} $
C. $\sqrt {\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{qQ}}{{m{R^3}}}} $
D. $\sqrt {\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{m}{{qQ}}} $

Answer 1

Hint:To solve this question, we will first consider the formula for the force acting between two charges. Then we will equate this force with the force due to Newton's second law of motion. We will use the concept of acceleration of a particle having SHM which includes angular frequency.

Formulas used:
$F = \dfrac{{kqQr}}{{{R^3}}}$
where, $F$ is the force between two charges, $Q$ and $q$ are the two charges, $r$ is the amplitude of the particle, $R$ is the radius of the sphere,
 $k = \dfrac{1}{{4\pi {\varepsilon _0}}}$
where ${\varepsilon _0}$ is the permittivity of the free space.
$F = ma$
where, $F$ is the force, $m$ is the mass of the particle and $a$ is the acceleration of the particle.
$a = - {\omega ^2}r$
where, $a$ is the acceleration of the particle, $\omega $ is the angular frequency of the particle and $r$ is the amplitude of the particle.

Complete step by step answer:
The force between the given two charges is given by the formula:
$F = \dfrac{{kqQr}}{{{R^3}}}$
Here we are given that the charge of the particle is $ - q$
$F = - \dfrac{{kqQr}}{{{R^3}}}$
Also, according to newton’s second law of motion, $F = ma$
Both these forces should be the same.
\[
ma = - \dfrac{{kqQr}}{{{R^3}}} \\
\Rightarrow a = - \dfrac{{kqQr}}{{m{R^3}}} \\ \]
But we know that the acceleration of a parting moving in SHM is given by,
$a = - {\omega ^2}r$.
$
\Rightarrow - {\omega ^2}r = - \dfrac{{kqQr}}{{m{R^3}}} \\
\Rightarrow {\omega ^2} = \dfrac{{kqQ}}{{m{R^3}}} \\ $
We know that $k = \dfrac{1}{{4\pi {\varepsilon _0}}}$
$
{\omega ^2} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{qQ}}{{m{R^3}}} \\
\therefore \omega = \sqrt {\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{qQ}}{{m{R^3}}}} \\
$
Thus, the angular frequency of the particle’s simple harmonic motion is $\sqrt {\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{qQ}}{{m{R^3}}}} $.

Hence, option C is the right answer.

Note:We have determined the value of the angular frequency of the particle having SHM. Simple harmonic motion is repetitive. The period $T$ is the time it takes the object to complete one oscillation and return to the starting position. The angular frequency ω is given by $\omega = \dfrac{{2\pi }}{T}$. The angular frequency is measured in radians per second.