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Show that a mass attached to a non ideal spring that has a restoring force given by $-kx-ax^3$, where $x$ is the displacement from the equilibrium position, executes a periodic motion. What's the period of this motion? Is it harmonic (simple harmonic motion)?

So, to show that the movement is periodic I guess I would need to solve the differential equation $mx''+kx+ax^3=0$ by Newton's second law. Is that right? I think there might be a better approach since this shouldn't be a difficult problem and should be solvable even for students that don't have a background on differential equations, but I can't get it right.

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  • $\begingroup$ Do you understand how to approach this using the conservation of energy rather than $F=ma$? $\endgroup$
    – Ghoster
    Commented Oct 2, 2023 at 19:40
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    $\begingroup$ multiply by x' and take a first integral $\endgroup$
    – trula
    Commented Oct 2, 2023 at 19:55

3 Answers 3

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Starting with the differential equation

$$m\,\ddot x+\,F(x)=0\tag 1$$

where $~\,F(x)=k\,x+a\,x^3~$

you can see the numerical solution with the initial conditions $~x(0)=x_0~,\dot x(0)=0~$ at this figure :

enter image description here

$~x(t)~$ is periodic motion but the period is depending on the amplitude of $~x(t)~$ which is $~x_0~$ , thus it is not simple harmonic motion

What's the period of this motion?

we want to obtain the time $~T~$ between the two blue points .

The period T

multiply equation (1) with $~\frac{dx}{dt}~$ and integrate you obtain the energy $~E=T+U~$

$$ E=\frac m2 \left(\frac{dx}{dt}\right)^2+U(x)\tag 2$$

where $~U(x)=\frac 12 k\,x^2+\frac 14 a\,x^4~$

at $~t=0~$ ,$~\frac{dx}{dt}=0~,x=x_0~$ thus $$E_0= \frac 12 k\,x_0^2+\frac 14 a\,x_0^4$$

solve the equation $~E=E_0~$ for $~dt~$ you obtain

$$dt=\pm 2\,{\frac {m{\it dx}}{\sqrt {-2\,m \left( 2\,{x}^{2}k+a{x}^{4}-2\,{x_ {{0}}}^{2}k-a{x_{{0}}}^{4} \right) }}} $$

from here the period $~T~$ is

$$T=\pm 4\,\int 2\,{\frac {m{\it dx}}{\sqrt {-2\,m \left( 2\,{x}^{2}k+a{x}^{4}-2\,{x_ {{0}}}^{2}k-a{x_{{0}}}^{4} \right) }}} $$

the result is "elliptic integral "

$$T=\pm 4\,\sqrt {m}{\it EllipticF} \left( {\frac {x}{x_{{0}}}},{\frac {ix_{{0 }}\sqrt {a}}{\sqrt {2\,k+a{x_{{0}}}^{2}}}} \right) \sqrt {2}{\frac {1} {\sqrt {2\,k+a{x_{{0}}}^{2}}}}\bigg|_{x=x0}^{x=0} $$

with the data of the above figure $~|T|=3.179~$


$$\dot x\,m\,\ddot x+\dot x \,F(x)=0\\ \frac m2 \frac{d}{dt} \left(\dot{x}^2\right)=-\frac {dx}{dt}\,F(x)\\ \frac m2 {d} \left(\dot{x}^2\right)=-{dx}\,F(x)\\ \frac m2 \int {d} \left(\dot{x}^2\right)=-\int {dx}\,F(x)$$

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Since no external forces are jerking off, the work is conservative and the energy remains the same for all instant of time. Hence, there must be a potential function $U$ which verify that $\vec{F} = - \vec{\nabla} U$.

With this sense, let $U(x) = \frac{1}{4} a x^4 + \frac{1}{2} k x^2$. Is not so difficult to verify that $U(x)$ represents a stable potential distribution since its concavity is positive. Therefore, $x$ does not diverge. Two options left: or $x$ is an oscillator or it reaches a steady value. Now, the total energy is:

$$E_0 = K(x) + U(x) = \frac{1}{2} m \left(\frac{dx}{dt} \right)^2 + \frac{1}{4} a x^4 + \frac{1}{2} k x^2$$

If $x$ reaches a steady value, this means that all the derivatives go to zero as t goes to infinity. The only possibility for this is that $x_\infty = 0$, otherwise $\sum \vec{F} \neq \vec{0}$. Replacing in the energy equation, we note that only the trivial solution with $E_0 = 0$ fulfills the condition. But we are interested in $E_0 > 0$, and then the system does not reach a steady value. But since $x(t)$ remains bounded, then $x(t)$ is periodic. The points of maximum elongation are where $x' = 0$:

$$E_0 = \frac{1}{4} a x^4 + \frac{1}{2} k x^2$$

$$\pm x_{MAX} = \pm \sqrt{\frac{\sqrt{4 E_0 a + k^2} - k}{a}} $$

The next step consists in finding $dt$ as a function of $x$, from the energy equation. Integrating from $-x_{MAX}$ to $x_{MAX}$ we get one half of the period:

$$dt = \sqrt{\frac{m}{2}} \frac{1}{\sqrt{E_0 - \frac{1}{4} a x^4 - \frac{1}{2} k x^2}} dx$$

$$ \int_0^{T/2} dt = \sqrt{\frac{m}{2}} \int_{-x_{MAX}}^{x_{MAX}} \frac{dx}{\sqrt{E_0 - \frac{1}{4} a x^4 - \frac{1}{2} k x^2}} $$

(good luck solving that)

Finally, the oscillator is not harmonic since the ODE is not linear.

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  • $\begingroup$ So I can find the potential function integrating the gradient of U(x) over x (there's only an horizontal component (depends on the reference, ofc)), right? if I integrate F over distance then I would get work, so the work done, in this case, is equal to the generated potential energy? Which means all the energy is converted from motion to potential until the body reaches x'=0 (max displacement). Is that reasoning correct? Thanks $\endgroup$ Commented Oct 6, 2023 at 15:35
  • $\begingroup$ In this case, the potential function is minus * (a 1D primitive of the force). Yo find one primitive of the force, which is equal to minus the gradient of U in the most general 3D case. So, if you integrate force over distance you get potential with the constant being the reference absolute potential you choose. The work is the change in potential energy between two points, is a ∆U $\endgroup$
    – Pizzaguy07
    Commented Oct 7, 2023 at 13:33
  • $\begingroup$ Hence, since the total energy remains constant because no external forces interfer, and since energy is K + U. And since the potential distribution U(x) is concave, the body has to reach a point where K=0 and then x'=0, assuming E distinct from zero $\endgroup$
    – Pizzaguy07
    Commented Oct 7, 2023 at 13:35
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Any time you have some acceleration defined as a function of position only you can always find the speed-distance relationship using the following direct integration

Given acceleration ${\rm acc}(x)$ omly

Subject to IC

$$ v(0) = v_0 $$

and

$$ x(0) = x_0 $$

First, integrate $ \underline{ v {\rm d}v = {\rm acc}(x) {\rm d}x }$ over distance

$$ \tfrac{1}{2} v^2 - \tfrac{1}{2} v_0^2 = \int \limits_{x_0}^x {\rm acc}(x) \,{\rm d}x $$

Then solve for $v$

$$ v = \sqrt{v_0^2 + 2 \int \limits_{x_0}^x {\rm acc}(x)\,{\rm d}x } $$

and possibly find the time-distance relationship with

$$t = \int \limits_{t_0}^t \frac{1}{v} {\rm d}x = \int \limits_{t_0}^t \frac{1}{\sqrt{v_0^2 + 2 \int \limits_{x_0}^x {\rm acc}(x)\,{\rm d}x }} {\rm d}x $$

For your case

$$ \tfrac{1}{2} v^2 - \tfrac{1}{2} v_0^2 = \int \limits_{x_0}^x \left( - \tfrac{k}{m} x - \tfrac{a}{m} x^3 \right) \,{\rm d}x = - \frac{ \tfrac{a}{2} (x^4-x_0^4) + k (x^2-x_0^2)}{2 m} $$

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