If a particle moves with acceleration = #10x-2x^3# and v= 0 at x=1, is the motion simple harmonic? And how do you describe the motion?

1 Answer
Jun 15, 2018

This is not SHM, because SHM requires a linear restorative force, ie SHM would require an acceleration in the form:

  • #a = - k x#

The potential function #V(x)# for this motion follows from:

#a = - (dV)/(dx) = 10x-2x^3#

#:. V(x) = -5x^2 + x^4/2 + V_o#

For #V_o = 0#, a plot of #V(x)# vs #x# looks like:

graph{-5x^2+x^4/2 [-5, 5, -15, 10]}

We are told that #v_((x=1)) = 0#.

Also:

  • #V(1) = - 4.5#

  • #V(3) = - 4.5#

The particle will thetefore oscillate back and forward in #x in [1,3]#.

This is a stable equilibrium.

Phase Portrait

#a = v (dv)/(dx) = 10x-2x^3#

Integrating wrt #x#:

#v^2/2 = 5x^2- x^4/2 + C#

#v(x = 1) = 4.5 + C = 0#

#v^2/2 = 5x^2- x^4/2 - 4.5#

#v = sqrt(10x^2- x^4 - 9)#

The graph of #v# vs #x# shows cycles on either side of the axis that correspond to this oscillation:

graph{y^2 = 10x^2- x^4 - 9 [-3.5, 3.5, -10, 10]}