Answer to Question #21238 in Mechanics | Relativity for swarup

Question

the potential energy of a particle of mass 1kg moving along x axis given by U(x)=[(X)sq/2 -x]joule .if total mechanical energy of particle is 2joule,find its maximum speed?

Answer 1

E=T+U
T=mx'^2/2
U=x^2/2-x
E=mx'^2/2+x^2/2-x
2=1*x'^2/2
Maximum speed is at point where x''=0
So, we have to solve next differential equation:
x'^2/2+x^2/2-x=2.
x'^2+x^2-2x=4
x'=Sqrt(-x^2+2x-4)
x''=(1-x)/Sqrt(-x^2+2x-4)
So, maximum we have at point x=1
At x=1 x'^2/2+1-2=4
x'^2/2=5
x'^2=10
x'=5=V - maximum speed