Oscillations and inverse square law

AI Thread Summary
The discussion revolves around finding the equilibrium position of a particle influenced by a constant force towards the origin and a repulsive inverse square law force. The initial attempt at a solution incorrectly simplified the forces involved, leading to an erroneous equation. The correct approach requires considering the sign of the forces, particularly the behavior of the inverse square force on either side of the origin. Participants emphasize the importance of solving for both positive and negative positions to accurately determine equilibrium. The conversation highlights the need for careful attention to sign conventions in physics problems.
Anchit
Messages
1
Reaction score
0

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?
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top