Understanding the Equation of Motion in Particle Mechanics: Explained

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The discussion centers on a particle mechanics problem involving a force that varies inversely with distance raised to the power of n. The equation of motion presented is d²x/dt² = -(u)/x^n, where u is a constant and x represents distance. Participants clarify that the symbol μ is simply an unknown constant and does not have specific physical significance. The equation is deemed to represent behavior in a theoretical context rather than actual physical phenomena. Overall, the focus is on understanding the mathematical formulation rather than its real-world application.
I am Meaningless
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Hello everybody, I Am Meaningless and I had this doubt when I was reading some illustrations based on questions in Particle mechanics.


So the Question was,

A particle is attracted by a force to a fixed point varying inversely as (distance)^n. If the velocity acquired in falling from an infinite distance to a distance 'a' from the centre be equal to the velocity acquired in falling from rest from distance 'a' to a distance (1/4)'a', prove that n = (3/2)
As this was an illustration it was natural for me to directly look at the solution given to me below it.
They started the problem like this:

The Equation of motion is

d^2x/dt^2 = -(u)/x^n
where u = meu (symbol) and x (from my observations) = distance covered

I did not understand this and would love it if someone could explain this equation

EDIT: I recognised the fact that
acceleration (inversely proportional) distance ^n
and when I equated would result in a proportionality constant, taken as 'u' and since it was being attracted had the negative symbol on it. But does the symbol have any significance? Any physical values?
 
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I am Meaningless said:
But does the symbol have any significance?
The symbol μ? No, it is just representing an unknown constant.
 
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haruspex said:
The symbol μ? No, it is just representing an unknown constant.
Thanks for your help, just was wondering if perhaps it really was an important equation.
 
I am Meaningless said:
Thanks for your help, just was wondering if perhaps it really was an important equation.
No, it represents the behaviour in this artificial question. It does not correspond to any actual physics of which I am aware.
 
haruspex said:
No, it represents the behaviour in this artificial question. It does not correspond to any actual physics of which I am aware.
Okay, Thanks for your help again![emoji1]
 
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