# 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?

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 $V \left(x\right)$ for this motion follows from:

$a = - \frac{\mathrm{dV}}{\mathrm{dx}} = 10 x - 2 {x}^{3}$

$\therefore V \left(x\right) = - 5 {x}^{2} + {x}^{4} / 2 + {V}_{o}$

For ${V}_{o} = 0$, a plot of $V \left(x\right)$ vs $x$ looks like:

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

We are told that ${v}_{\left(x = 1\right)} = 0$.

Also:

• $V \left(1\right) = - 4.5$

• $V \left(3\right) = - 4.5$

The particle will thetefore oscillate back and forward in $x \in \left[1 , 3\right]$.

This is a stable equilibrium.

Phase Portrait

$a = v \frac{\mathrm{dv}}{\mathrm{dx}} = 10 x - 2 {x}^{3}$

Integrating wrt $x$:

${v}^{2} / 2 = 5 {x}^{2} - {x}^{4} / 2 + C$

$v \left(x = 1\right) = 4.5 + C = 0$

${v}^{2} / 2 = 5 {x}^{2} - {x}^{4} / 2 - 4.5$

$v = \sqrt{10 {x}^{2} - {x}^{4} - 9}$

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]}