Question

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?

Answer 1

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]}