Sunday, 9 September 2018

homework and exercises - Finding the equation of motion of anharmonic potential



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[4xa24x3a4]
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[12(xa)2+(xa)4]


I found that the value in brackets reduces to 1a2(x2a2)2



So: dxdt=2U0mx2a2a2

So I want xx0dxx2a2=t02U0mdta2
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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