# Oscillations and inverse square law

## Homework Statement

A particle of mass m moves in 1 dimension along positive x direction.It is acted on by a constant force directed towards origin with magnitude B,and an inverse square law repulsive force with magnitude A/x^2.Find equilibrium position.

B+A/x^2=m*a

## The Attempt at a Solution

B+A/x^2=0
x=sqrt(-A/B)

Is this correct?

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ShayanJ
Gold Member
No, the correct form is:
$m\ddot x=\frac{A}{x^2}-B sign(x)$

Where $sign(x)=\left\{ \begin{array}{ll} 1 & x>0 \\ 0 & x=0 \\ -1 & x<0 .\end{array} \right.$

Then you should solve for x<0 and x>0 separately. I hope the particle won't pass the origin!

jbriggs444
Homework Helper
2019 Award
No, the correct form is:
$m\ddot x=\frac{A}{x^2}-B sign(x)$
Shyan's equation ignores the fact that the inverse square force also changes sign on opposite sides of the origin. If you include that fact, the sign() function can be simplified away.

Anchit's equation ignores the sign convention assumed in the problem: Both A and B are positive there. We do not want to be taking the square root of a negative number!!

Heed Shyan's advice and look for both positive and negative solutions.