If I have a potential given by: U=U0[2(xa)2−(xa)4]
It says that at t=0, the particle is at the origin (x=0) and the velocity is positive and equal to the escape velocity, which I found to be √2U0/m I have the differential equation: m¨x=−∇U=−U0[4xa2−4x3a4]
EDIT::
So I have the initial (x,˙x). Now to find x(t), I use conservation of energy.
E=K+U=12mv2+U0[2(xa)2−(xa)4]
The energy of the system is U0, so I can change the above equation to: U0=12m(dxdt)2+U0[2(xa)2−(xa)4]
(dxdt)2=2mU0[1−2(xa)2+(xa)4]
I found that the value in brackets reduces to 1a2(x2−a2)2
So: dxdt=√2U0mx2−a2a2
So I want ∫xx0dxx2−a2=∫t0√2U0mdta2
This ends up being a hyperbolic arctan function, which could potentially make sense, but am I going in the right direction?
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