Oscillations and inverse square law

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SUMMARY

The discussion focuses on the dynamics of a particle influenced by a constant force and an inverse square law repulsive force. The correct equation of motion is established as m\ddot x=\frac{A}{x^2}-B sign(x), which accounts for the directionality of the forces involved. Participants emphasize the importance of considering both positive and negative solutions for the equilibrium position, specifically noting that the equilibrium position is given by x=sqrt(-A/B) under the correct conditions. Misinterpretations regarding the sign of forces and the nature of the inverse square law are clarified.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with inverse square law forces
  • Knowledge of differential equations
  • Concept of equilibrium in mechanics
NEXT STEPS
  • Study the derivation of equations of motion for forces acting on particles
  • Explore the implications of the inverse square law in different physical contexts
  • Learn about the sign function and its applications in physics
  • Investigate stability analysis for equilibrium positions in mechanical systems
USEFUL FOR

Students of physics, particularly those studying classical mechanics, as well as educators and anyone interested in the mathematical modeling of forces and motion.

Anchit
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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.

Homework Equations



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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No, the correct form is:
<br /> m\ddot x=\frac{A}{x^2}-B sign(x)<br />

Where sign(x)=\left\{ \begin{array}{ll} 1 &amp; x&gt;0 \\ 0 &amp; x=0 \\ -1 &amp; x&lt;0 .\end{array} \right.

Then you should solve for x<0 and x>0 separately. I hope the particle won't pass the origin!
 
Shyan said:
No, the correct form is:
<br /> m\ddot x=\frac{A}{x^2}-B sign(x)<br />

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.
 

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