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A particle moves with force

$$F(x) = -kx +\frac{kx^3}{A^2}$$

Where k and A are positive constants.

if $KE_o$ at x = 0 is $T_0$ what is the total energy of the system?

$$ \Delta\ KE(x) + \Delta\ U(x) = 0$$

$$F(x) = -\frac{dU}{dx} = m\frac{dv}{dt} = m v\frac{dv}{dx}$$

Integrating to get U(x) and 1/2mv^2 I get

$$\Delta\ U(x) = \frac{kx^2}{2} - \frac{kx^4}{4A^2}$$

$$\Delta\ KE(x) = -\frac{kx^2}{2} + \frac{kx^4}{4A^2}$$

Which Makes sense. But how do I find the function KE(x) where KE(0) = $T_0$? Do I Even need to? The total energy in the system is $T_0$ Correct?

Also a kind of side note. What is really confusing me, is when should I add limits of integration and under what circumstances should I just use an indefinite Integral?

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2 Answers 2

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The question is a little tricky as stated... Because your force is conservative, it can be written as the gradient of a scalar potential field. But the potential field, i.e. the potential energy, is defined only up to a constant. That is, your potential energy field is

$$U(x) = \frac{kx^2}{2}-\frac{kx^4}{4A^2}+U_0,$$

for any value of $U_0$. This U_0 comes up because we have done an indefinite integral of the force field to find it. So in all purity, the total energy of your particle is $T_0+U_0$, so it can be whatever you want, because you can choose $U_0$ freely...

For gravitational potential the convention is to place zero energy at infinity, for potentials such as yours it makes more sense to have zero energy at the origin. This is equivalent to choosing $U_0=0$. Which is a very reasonable thing to do, but in no way mandatory.

The full definite integral thing to arrive at conservation of energy from $F = ma$ is as follows,

$$ma = F(x)$$ $$m\frac{dv}{dx}v=F(x)$$ $$mvdv = F(x)dx$$ $$\int_{v_0}^{v_1}{mvdv}=\int_{x_0}^{x_1}{F(x)dx}$$ $$\left.\frac{1}{2}mv^2\right|_{v_0}^{v_1} = \left. (-\frac{kx^2}{2}+\frac{kx^4}{4A^2})\right|_{x_0}^{x_1}$$ $$\frac{1}{2}mv_1^2 - \frac{1}{2}mv_0^2 = -\frac{kx_1^2}{2}+\frac{kx_1^4}{4A^2} + \frac{kx_0^2}{2}-\frac{kx_0^4}{4A^2}$$ $$\frac{1}{2}mv_1^2 +\frac{kx_1^2}{2}-\frac{kx_1^4}{4A^2} = \frac{1}{2}mv_0^2+ \frac{kx_0^2}{2}-\frac{kx_0^4}{4A^2}$$ $$T_1 +U_1 = T_0+U_0$$

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For this particular problem, it is useful to note that the force function has three zeros:

$$F(0) = F(-A) = F(A) = 0$$

This means the potential has three stationary points. Looking at the potential function, we see that $U(0)$ is a local minimum while $U(-A) = U(A)$ are global maximums.

This suggests that a natural choice is to set $U(0) = 0$.

$$U(x) = \frac{kx^2}{2} - \frac{kx^4}{4A^2}$$

Thus the total energy is equal to the kinetic energy when $x=0$:

$$E = T_0$$

For $T_0 < kA^2/4$, the particle is bound by the potential and just oscillates back and forth symmetrically about $x=0$, never reaching $|x| = A$.

For $T_0 > kA^2/4$, the particle is unbound and is driven to plus or minus infinity by the potential (the particle rolls down the potential hill on either side).

For $T_0 = kA^2/4$, the particle theoretically would stop at either $x=-A$ or $x=A$, depending on the initial conditions, and remain there. However, both points are unstable, i.e., an infinitesimal perturbation would send the particle back towards $x=0$ or towards plus or minus infinity.

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