# Differential equation for the anharmonic oscillator

In my project me and my partner used the engine to constrain the system so we can see the anharmonic oscillations. In our first analysis we get only odd powers in differential equation, so there should be only odd harmonic oscilations ($$\omega$$, $$3\omega$$ etc.). But in the real data we also get the second harmonic oscilations ($$2\omega$$). So we think that if we can do the Taylor series about the equilibrium and not the zero we can get in the equation a second order (the eqaution is below) of $$x$$ so there would be the second harmony.

In the picture you can see the system: there is an engine on a floor that moves the weight. $$l_0$$ is the initial length of the up springs, $$l_{02}$$ is the initial lentgh of the down spring, $$\Deltal$$ is a length change of an up spring, $$k_1$$ is the constant of both up springs, $$k_2$$ is the constant of a down spring which connects the system with the engine, $$a$$ is the distance between the path and the collumns, $$x$$ is our coordinate, $$x_e$$ is the coordinate of the engine's spring.

$$x_e(t)=A\sin(\omega t)$$ $$l(t)=\sqrt{x^2+a^2}$$ $$\Delta l = l(t)-l_0 = \sqrt{x^2+a^2} - l_0$$ $$F_{k1} = -k_1\Delta l \sin(\theta);\ \sin(\theta)=\frac{x}{\sqrt{x^2+a^2}}$$ $$F_{k2} = k_2(x-x_e+l_{02})$$ So the equation is: $$m\ddot{x}=2F_{k1}-F_{k2}-mg$$ We define new constants: $$\omega _1=\sqrt\frac {2k_1}{m};\ \omega _2=\sqrt\frac {k_2}{m}$$ And so we have: $$\ddot{x}+\omega _1 ^2 x(1-\frac{l_0}{\sqrt{x^2+a^2}})+ \omega _2 ^2 x=- \omega _2 ^2 (l_{02}-x_e)-g$$ As you can see if we do the Taylor series about $$0$$ for the square root, we can get only even powers since $$x^2$$ is under the root. So if we product it with $$x$$, we get the odd powers. Therefore there shouldn't be even harmonic oscilations. If we take the $$\ddot x=0$$ to find the equilibrium(s), we get the equation that after some mathematics gets the look of 4th order: $$(\omega _1 ^2 +\omega _2 ^2)x^4 +2(\omega _1 ^2 +\omega _2 ^2)[g+ \omega _2 ^2 (x_e-l_{02})]x^3+ [(\omega _1 ^2 +\omega _2 ^2)a^2+[g+ \omega _2 ^2 (x_e-l_{02})]^2-l_0 ^2 \omega _1 ^2]x^2+ 2(\omega _1 ^2 +\omega _2 ^2)[g+ \omega _2 ^2 (x_e-l_{02})]a^2 x +[g+ \omega _2 ^2 (x_e-l_{02})]^2=0$$

I use the Matlab to find the roots of this equation and have 4 roots. Each of them takes 37 pages A4 of Word in Arial 12. It is difficult to work with these solutions and understand which of them is the equilibrium we need. Is there any other way to find the equilibriums? Or is there another way we can find how the second order of $$x$$ enters to the equation?

you want to find x (equilibrium) that fulfilled this eqaution.

$$f(x)=\omega_{{1}}x \left( 1-{\frac {l_{{0}}}{\sqrt {{x}^{2}+{a}^{2}}}} \right) +{\omega_{{2}}}^{2}x+{\omega_{{2}}}^{2} \left( l_{{2}}-x_{{E} } \right) +g=0\tag 1$$

first take the Taylor series of

$${\frac {l_{{0}}}{\sqrt {{x}^{2}+{a}^{2}}}}\approx{\frac {l_{{0}}}{\sqrt {{a}^{2}}}}-\frac 12\,{\frac {l_{{0}}{x}^{2}}{\sqrt {{a}^{2}}{a}^{2}}}$$

put it in Eq. (1)

$$f(x)\mapsto a_0+a_1\,x+a_3\,x^3=0\tag 2$$

with: $$a_0={\omega_{{2}}}^{2} \left( l_{{2}}-x_{{E}} \right) +g$$ $$a_1=\omega_{{1}} \left( 1-{\frac {l_{{0}}}{\sqrt {{a}^{2}}}} \right) +{ \omega_{{2}}}^{2}$$ $$a_3=1/2\,{\frac {\omega_{{1}}l_{{0}}}{\sqrt {{a}^{2}}{a}^{2}}}$$

you have three solutions of Eq. (2) but just one is real value

$$x_{\text{real}}= \frac 16\,\sqrt [3]{ \left( -108\,{\it a_0}+12\,\sqrt {3}\sqrt {{\frac {4\,{ {\it a_1}}^{3}+27\,{{\it a_0}}^{2}{\it a_3}}{{\it a_3}}}} \right) {{\it a_3 }}^{2}}{{\it a_3}}^{-1}-2\,{\it a_1}{\frac {1}{\sqrt [3]{ \left( -108\,{ \it a_0}+12\,\sqrt {3}\sqrt {{\frac {4\,{{\it a_1}}^{3}+27\,{{\it a_0}}^{ 2}{\it a_3}}{{\it a_3}}}} \right) {{\it a_3}}^{2}}}} \tag 3$$

to valid the solution of Eq. (3) I put some data

$$[\omega_{{1}}=10,\omega_{{2}}=2,g=10,l_{{0}}= 0.9,l_{{2}}= 0.3,x_{{E}} = 2.5,a=3] ~$$ and got $$x_\text{real}=-0.109~$$ if you use this data to find the solution of Eq. (1) you get the same result , so at list for those data is the ansatz correct.

I use MAPLE to do the symbolic results

• thank you, it is very helpful Jan 22, 2021 at 13:03