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A particle of mass m is located in a uni...

A particle of mass `m` is located in a unidimensional potential field where potential energy of the particle depends on the coordinates `x as U (x) = (A)/(x^(2)) - (B)/(x)` where `A` and `B` are positive constant.
Find the time period of small oscillation that the particle perform about equilibrium position.

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Verified by Experts

The correct Answer is:
`4 pi A sqrt((mA)/(B^(4)))`

The potential energy friction for the mass m in a unidimensional potential field is given as:
`U (x) = (A)/(x^(2)) - (B)/(x)` (i)
Since this is the potential energy of mass m is a potential field, so the force acting on the mass is obtained by the relation:
`F = - (dU)/(dx) = (2 A)/(x^(3)) - (B)/(x^(2))`
Now, the equilibrium position is given by
`F = 0 or (2 A)/(x^(3)) - (B)/(x^(2)) = 0`
`x^(2) (2A - Bx) = 0`
So, `x = 0` or `x = (2A)/(B)`
Since `x = 2 A//B` gives more stable situation
(because energy is minimum here : `d^(2) U// dx^(2)` is positive). slabe equilibrum position is given as `x = (2 A)/(B)`
Now if the mass is displaced slightly away from the equilibrium position, the force acting on mass m is given as
`F = (2 A)/[(2A//B) + Delta x]^(2) - (B)/[(2A//B) + Delta x]^(2)`
`= (2 A)/(((2 A)/(B))^(3) (1 + (B Delta x)/(2 A))^(2)) - (B)/(((2 A)/(B))^(2) (1 + (B Delta x)/(2 A))^(2))`
`= (2 A)/(8A^(2)//B^(3) [1+ (B Delta x)/(2 A)]^(-3)) - (B)/(4A^(2)//B^(2) [1+ (B Delta x)/(2 A)]^(-2))`
So, the bionomial expansion for small `Deltax`, we get
`F = (B^(3))/(4 A^(2)) [1- (B Delta x)/(2 A)] - (B^(3))/(4 A^(2)) [1- (2 B Delta x)/(2 A)]`
`= (B^(3))/(4 A^(2)) [1- (3 B Delta x)/(2 A)] - 1 + [1+ (2 B Delta x)/(2 A)] = - ((B^(4))/(8 A^(3))) Delta x`
We can write
`m (d^(2)x)/(dt^(2)) = - (B^(4))/(8 A^(3)) Delta x`
`(d^(2)x)/(dt^(2)) = - ((B^(4))/(8 A^(3))) Delta x` (iii)
`(d^(2)x)/(dt^(2)) prop - Delta x`
So, from the relations, it is clear that partical will perform SHM. But for any SHM.
`(d^(2)x)/(dt^(2)) = - omega^(2) Delta x`
So, from Eqs. (iii) and (iv), we have
`omega^(2) = (B^(4))/(8A^(2)m)`
`T = (2 pi)/(omega) = 2 pi sqrt((8A^(2)m)/(B^(4))) = 4 pi A sqrt((mA)/(B^(4)))`
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