Question

The acceleration of a particle moving along the x-axis is given by a = `2x^3-10x`. a) if u = 3 show that the particle oscillates within the interval `-1<=x<=1` b) Why is it not simple harmonic motion c) if u=6 describe the motion?

Answer 1

See below

Explanation:

NB: The full question is actually in the comments, some of it repeated here.

If the particle starts at the origin with velocity u, then its velocity v is given by:

• `v^2-u^2=x^4-10x^2 qquad square`

(a)

If `u=3`, show that the particle oscillates within the interval:

• `-1 lt =x lt =1`

From `square`, with `u = 3`:

`v(x) = sqrt( x^4 - 10x^2 + 9 )`, plotted below

graph{sqrt( x^4 - 10x^2 + 9 ) [-10, 10, -5, 5]}

• `v = 0, qquad x = -3,-1,1,0`

But from the plot, the function disappears in `[-3,-1]` and `[1,3]` because: `x^4 - 10x^2 + 9 < 0`.

This is a very unsatisfactory way to be looking at physical problems.

Re-writing `square`:

• `underbrace(1/2v^2)_("KE") + underbrace(5x^2 - 1/2x^4)\_("PE") = underbrace(1/2u^2)\_("TE") qquad triangle`

This is now an expression for the (unit mass) Kinetic and Potential Energies of the system, the Total Energy being the constant: `1/2 u^2`

For `u = 3`, `TE = 4.5`.

• `{(x = -1, KE = 0, PE = 4.5),(x = 0, KE = 4.5, PE = 0),(x - 1, KE = 0, PE = 4.5):}`

Most revealing is the potential well :

• `PE(x) = 5x^2 - 1/2x^4`

graph{5x^2 - 1/2x^4 [-4, 4, -5, 20]}

The particle will oscillate in `[-1,1]`. It does not have enough energy to rise to the peaks at `x = - sqrt5, sqrt5`:

• `PE(- sqrt5) = PE(sqrt5) = 12.5`

(b)

Short answer: SHM is premised on there being a linear restorative force:

• `F = - kx`

You could linearise the motion here about the origin and obtain approximate SHM like with the simple pendulum.

(c)

For `u = 6`, `TE = 18 bb( gt 12.5)`.

The motion is unstable . The particle will climb out of the well and accelerate to the right.

Total Energy will remain constant, but once the perticle scales the peak constantly PE will be traded for KE.