Quartic Oscillator: Solving for Time T to Reach Max Amplitude

[ZoroP]

Homework Statement

The equation of motion for a particle of mass 1 in a quartic oscillator V(x)=0.25x^4 is x''+x^3=0. Suppose that the maximum amplitude of the oscillator is Xm(max). Find an expression for the time T that it takes to go from x=0 to x=Xm(max) and show that this time is inversely proportional to Xm(max).

Homework Equations

Xm(max) = sqrt(x^2+mv^2/k)

The Attempt at a Solution

V(x) is energy? E = 0.5mv^2 + V then d^2/dt^2(x) = -x^3 ? How to make the connection between the time T and these equations. Thanks very much.

 

[tiny-tim]

Hi ZoroP!

(try using the X² tag just above the Reply box )

The equation of motion for a particle of mass 1 in a quartic oscillator V(x)=0.25x^4 is x''+x^3=0.
…
V(x) is energy? …

Hint: multiply by x' to give x'x'' + x³x' = 0, and then integrate.

 

[ZoroP]

Thx tiny, then i get v*dv/dt + x^3 dx/dt = 0
how to integral this? is this the same with v*dv = - x^3 dx?

 

[tiny-tim]

Hi ZoroP!

(just got up … :zzz:)

(please use the X² tag just above the Reply box )

Thx tiny, then i get v*dv/dt + x^3 dx/dt = 0
how to integral this? is this the same with v*dv = - x^3 dx?

Yes, they're the same …

so what do you get? ​