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A particle rests in equilibrium under two forces of repulsion whose centres are at a distance of a and b from the particle.

Question :- A particle rests in equilibrium under two forces of repulsion whose centres are at a distance of $a$ and $b$ from the particle. The forces vary as the cube of the distance. The forces per unit mass are $k$ and $k^{'}$ respectively. If the particle is slightly displaced towards one of them then, the motion is simple harmonic with the time period equal to:

(a) $\displaystyle \frac{2\pi }{\sqrt{3\left( \frac{k}{{{a}^{3}}}+\frac{k'}{{{b}^{3}}} \right)}}$

(b) $\displaystyle \frac{2\pi }{\sqrt{\left( \frac{k}{{{a}^{3}}}+\frac{k'}{{{b}^{3}}} \right)}}$

(d) $\displaystyle \frac{2\pi }{\sqrt{3\left( \frac{k}{{{a}^{4}}}+\frac{k'}{{{b}^{4}}} \right)}}$

Solution :-

Let the mass of the particle is ‘m’.

As the forces vary as “cube of distance” and the force per unit mass are k and k’ so the forces can be written as :

$\displaystyle {{F}_{1}}=\frac{mk}{{{a}^{3}}}$

and

$\displaystyle {{F}_{2}}=\frac{mk'}{{{b}^{3}}}$

Since initially the particle is at rest, so

F₁ = F₂

$\displaystyle \Rightarrow \frac{k}{{{a}^{3}}}=\frac{k'}{{{b}^{3}}}$ …..(1)

Now let the particle is slightly(let = x) displaced towards on the agent applying force, then net force on it becomes :

$\displaystyle {{F}_{net}}=\frac{mk}{{{(a-x)}^{3}}}-\frac{mk'}{{{(b+x)}^{3}}}$

$\displaystyle \Rightarrow {{F}_{net}}=m\left[ \frac{k}{{{a}^{3}}{{\left( 1-\frac{x}{a} \right)}^{3}}}-\frac{k'}{{{b}^{3}}{{\left( 1+\frac{x}{b} \right)}^{3}}} \right]$

$\displaystyle \Rightarrow {{F}_{net}}=m\left[ \frac{k}{{{a}^{3}}}{{\left( 1-\frac{x}{a} \right)}^{-3}}-\frac{k'}{{{b}^{3}}}{{\left( 1+\frac{x}{b} \right)}^{-3}} \right]$

Expanding using binomial theorem and neglecting higher terms,

$\displaystyle \Rightarrow {{F}_{net}}=m\left[ \frac{k}{{{a}^{3}}}\left( 1+\frac{3x}{a} \right)-\frac{k'}{{{b}^{3}}}\left( 1-\frac{3x}{b} \right) \right]=m\left[ \frac{k}{{{a}^{3}}}+\frac{3kx}{{{a}^{4}}}-\frac{k'}{{{b}^{3}}}+\frac{3k'x}{{{b}^{4}}} \right]$

From equation (1) as $\displaystyle \Rightarrow \frac{k}{{{a}^{3}}}=\frac{k'}{{{b}^{3}}}$ , so we get

$\displaystyle {{F}_{net}}=3mx\left[ \frac{k}{{{a}^{4}}}+\frac{k'}{{{b}^{4}}} \right]$

Now here this net force is actually restoring force which is trying to pull back the body to its initial position, so applying a negative sign, we get

$\displaystyle {{F}_{restoring}}=-3mx\left[ \frac{k}{{{a}^{4}}}+\frac{k'}{{{b}^{4}}} \right]$

Comparing the above equation with F = -kx, we get

$\displaystyle k=3m\left[ \frac{k}{{{a}^{4}}}+\frac{k'}{{{b}^{4}}} \right]$

Now

$\displaystyle {{\omega }^{2}}=\frac{k}{m}=3\left[ \frac{k}{{{a}^{4}}}+\frac{k'}{{{b}^{4}}} \right]$

$\displaystyle \omega =\sqrt{3\left[ \frac{k}{{{a}^{4}}}+\frac{k'}{{{b}^{4}}} \right]}$

$\displaystyle T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{3\left[ \frac{k}{{{a}^{4}}}+\frac{k'}{{{b}^{4}}} \right]}}$

A particle rests in equilibrium under two forces of repulsion whose centres are at a distance of a and b from the particle.

A particle rests in equilibrium under two forces of repulsion whose centres are at a distance of a and b from the particle.

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