# Autoparametric resonance

<center></center>

System of equations of motion for the system:
\begin{aligned}
\ddot{R}&=R\dot{\theta}^2-\frac{k}{m}(R-l)+g\cos\theta+a\cos\Omega t\\
\ddot{\theta}&=-\frac{2}{R}\dot{R}\dot{\theta}-\frac{g}{m}\sin\theta
\end{aligned}

Find stationary point for linearization $\theta=0$, all derivatives are zero:
$$-\frac{k}{m}(R-l)+g=0\quad\implies\quad R=l+\frac{mg}{k}$$
linearize equations around stable point
$$R=R_0+r,\quad R_0=l+mg/k,\quad\theta_0=0$$
where $r/R_0\ll0$ and $\theta\ll0$. Scale time to get rid of frequencies $t'=\omega_1t$ assuming $\omega_1=2\omega_2$.
\begin{aligned}
\ddot{r}+\frac{1}{Q_1}\dot{r}+r&=l\dot{\theta}^2+a\cos\Omega t\\
\ddot{\theta}+\frac{1}{Q_2}\dot{\theta}+\frac{1}{4}\theta&=-2\dot{r}\dot{\theta}
\end{aligned}

Following to the averaging method:
\begin{aligned}
r &= u_1\cos\Omega t+v_1\sin\Omega t\\
\theta &= u_2\cos\frac{\Omega}{2}t+v_2\sin\frac{\Omega}{2}t\\
\dot{r} &= \Omega (-u_1\sin\Omega t+v_1\cos\Omega t)\\
\dot{\theta} &= \frac{\Omega}{2} (-u_2\sin\frac{\Omega}{2}t+v_2\cos\frac{\Omega}{2}t)
\end{aligned}
with additional constraints
\begin{aligned}
\dot{u}_1\cos\Omega t+\dot{v}_1\sin\Omega t &= 0\\
\dot{u}_2\cos\frac{\Omega}{2}t+\dot{v}_2\sin\frac{\Omega}{2}t &= 0
\end{aligned}

Introduce detuning parameters
$$\xi_1=1-\Omega^2,\quad\xi_2=1-\Omega^2/4$$


```python
import sympy as sp
```


```python
t, l, a, Q1, Q2 = sp.symbols('t, l, a, Q_1, Q_2')
u1, u2, v1, v2 = sp.symbols('u_1, u_2, v_1, v_2')
xi1, xi2 = sp.symbols('\\xi_1, \\xi_2')
W = sp.Symbol('\Omega', positive=True)
```


```python
r  =       u1 * sp.cos(W*t) + v1 * sp.sin(W*t)
dr = W * (-u1 * sp.sin(W*t) + v1 * sp.cos(W*t))
```


```python
th  =         u2 * sp.cos(W/2*t) + v2 * sp.sin(W/2*t)
dth = W/2 * (-u2 * sp.sin(W/2*t) + v2 * sp.cos(W/2*t))
```


```python
F1 = -dr/Q1/W  - xi1/W * r + l/W * dth**2 + a/W *sp.cos(W*t)
F2 = -dth/Q2/W - xi2/W * th - 2/l/(W/2) * dr * dth
```


```python
T = 4*sp.pi/W
```


```python
du1 = (sp.integrate(-sp.sin(W*t)*F1, (t, 0, T))/T).expand()
dv1 = (sp.integrate( sp.cos(W*t)*F1, (t, 0, T))/T).expand()
```


```python
du2 = (sp.integrate(-sp.sin(W/2*t)*F2, (t, 0, T))/T).expand()
dv2 = (sp.integrate( sp.cos(W/2*t)*F2, (t, 0, T))/T).expand()
```

Get equations for the steady-state solution.


```python
eq1=du1*2
eq1
```




$\displaystyle \frac{\Omega l u_{2} v_{2}}{4} + \frac{\xi_{1} v_{1}}{\Omega} - \frac{u_{1}}{Q_{1}}$




```python
eq2=dv1*2
eq2
```




$\displaystyle - \frac{\Omega l u_{2}^{2}}{8} + \frac{\Omega l v_{2}^{2}}{8} - \frac{\xi_{1} u_{1}}{\Omega} + \frac{a}{\Omega} - \frac{v_{1}}{Q_{1}}$




```python
eq3 = du2*2
eq3
```




$\displaystyle - \frac{\Omega u_{1} v_{2}}{l} + \frac{\Omega u_{2} v_{1}}{l} + \frac{\xi_{2} v_{2}}{\Omega} - \frac{u_{2}}{2 Q_{2}}$




```python
eq4=dv2*2
eq4
```




$\displaystyle - \frac{\Omega u_{1} u_{2}}{l} - \frac{\Omega v_{1} v_{2}}{l} - \frac{\xi_{2} u_{2}}{\Omega} - \frac{v_{2}}{2 Q_{2}}$



Replace real $u$, $v$ with complex analogs to easy account for damping
$$z_1=u_1+iv_1,\quad z_2=u_1-iv_1,\quad z_3=u_2+iv_2,\quad z_4=u_2-iv_2$$


```python
z1, z2, z3, z4=sp.symbols('z_1, z_2, z_3, z_4')
```


```python
sbs=[(u1, z1/2+z2/2), (v1, -sp.I*z1/2+sp.I*z2/2),
     (u2, z3/2+z4/2), (v2, -sp.I*z3/2+sp.I*z4/2)]
```


```python
eq5=eq1.subs(sbs).expand()
eq6=eq2.subs(sbs).expand()
eq7=eq3.subs(sbs).expand()
eq8=eq4.subs(sbs).expand()
```


```python
eq9  = (eq5 + sp.I * eq6).expand()
eq10 = (eq5 - sp.I * eq6).expand()
eq11 = (eq7 + sp.I * eq8).expand()
eq12 = (eq7 - sp.I * eq8).expand()
```


```python
eq9
```




$\displaystyle - \frac{i \Omega l z_{3}^{2}}{8} - \frac{i \xi_{1} z_{1}}{\Omega} + \frac{i a}{\Omega} - \frac{z_{1}}{Q_{1}}$




```python
eq10
```




$\displaystyle \frac{i \Omega l z_{4}^{2}}{8} + \frac{i \xi_{1} z_{2}}{\Omega} - \frac{i a}{\Omega} - \frac{z_{2}}{Q_{1}}$




```python
eq11
```




$\displaystyle - \frac{i \Omega z_{1} z_{4}}{l} - \frac{i \xi_{2} z_{3}}{\Omega} - \frac{z_{3}}{2 Q_{2}}$




```python
eq12
```




$\displaystyle \frac{i \Omega z_{2} z_{3}}{l} + \frac{i \xi_{2} z_{4}}{\Omega} - \frac{z_{4}}{2 Q_{2}}$




```python
sol1 = sp.solve((eq9,eq10),[z1,z2])
Z1 = sol1[z1].expand()
Z2 = sol1[z2].expand()
```


```python
Z1
```




$\displaystyle \frac{i Q_{1} \Omega^{2} l z_{3}^{2}}{- 8 i Q_{1} \xi_{1} - 8 \Omega} - \frac{8 i Q_{1} a}{- 8 i Q_{1} \xi_{1} - 8 \Omega}$




```python
Z2
```




$\displaystyle - \frac{i Q_{1} \Omega^{2} l z_{4}^{2}}{8 i Q_{1} \xi_{1} - 8 \Omega} + \frac{8 i Q_{1} a}{8 i Q_{1} \xi_{1} - 8 \Omega}$



Substitute them into the second couple of equations


```python
eq13 = eq11.subs([(z1,Z1),(z2,Z2)]).expand()
eq13
```




$\displaystyle \frac{Q_{1} \Omega^{3} z_{3}^{2} z_{4}}{- 8 i Q_{1} \xi_{1} - 8 \Omega} - \frac{8 Q_{1} \Omega a z_{4}}{- 8 i Q_{1} \xi_{1} l - 8 \Omega l} - \frac{i \xi_{2} z_{3}}{\Omega} - \frac{z_{3}}{2 Q_{2}}$




```python
eq14 = eq12.subs([(z1,Z1),(z2,Z2)]).expand()
eq14
```




$\displaystyle \frac{Q_{1} \Omega^{3} z_{3} z_{4}^{2}}{8 i Q_{1} \xi_{1} - 8 \Omega} - \frac{8 Q_{1} \Omega a z_{3}}{8 i Q_{1} \xi_{1} l - 8 \Omega l} + \frac{i \xi_{2} z_{4}}{\Omega} - \frac{z_{4}}{2 Q_{2}}$



Let the amplitudes of modes are
$$A_1^2 = z_1z_2 = u_1^2+v_1^2,\quad A_2^2 = z_3z_4 = u_2^2+v_2^2$$
The trivial solution to the system is $z_3=z_4=0$, $A_2=0$ and
\begin{equation}
z_{1,2}=\frac{a}{\pm\xi_1-\Omega/{iQ_1}},\quad A_1=\frac{a}{\sqrt{(1-\Omega^2)^2+\Omega^2/Q_1^2}}
\end{equation}

The system has nontrivial value


```python
A1, A2 = sp.symbols('A1, A2')
```


```python
eq15 = Q1*W**3*z3*A2**2/(-8*sp.I*Q1*xi1 - 8*W) - 8*Q1*W*a*z4/(-8*sp.I*Q1*xi1*l - 8*W*l) - sp.I*xi2*z3/W - z3/(2*Q2)
eq16 = Q1*W**3*z4*A2**2/(8*sp.I*Q1*xi1 - 8*W) - 8*Q1*W*a*z3/(8*sp.I*Q1*xi1*l - 8*W*l) + sp.I*xi2*z4/W - z4/(2*Q2)
```


```python
eq17 = eq15.collect([z3,z4])
eq18 = eq16.collect([z3,z4])
```

The linear system of two equations for $z_3$ and $z_4$ has a nontrivial solution when determinant is zero what yields $A_2^2$ value. Since A_2 is amplitude, nontrivial solution will be stable when $A_2$ has a real positive value, otherwise the solution is trivial.


```python
eq17
```




$\displaystyle - \frac{8 Q_{1} \Omega a z_{4}}{- 8 i Q_{1} \xi_{1} l - 8 \Omega l} + z_{3} \left(\frac{A_{2}^{2} Q_{1} \Omega^{3}}{- 8 i Q_{1} \xi_{1} - 8 \Omega} - \frac{i \xi_{2}}{\Omega} - \frac{1}{2 Q_{2}}\right)$




```python
eq18
```




$\displaystyle - \frac{8 Q_{1} \Omega a z_{3}}{8 i Q_{1} \xi_{1} l - 8 \Omega l} + z_{4} \left(\frac{A_{2}^{2} Q_{1} \Omega^{3}}{8 i Q_{1} \xi_{1} - 8 \Omega} + \frac{i \xi_{2}}{\Omega} - \frac{1}{2 Q_{2}}\right)$




```python

```

6f4a87247ac2987d9639bf2e7ac8e54a4aa872e7
ipynb
Jupyter Notebook
autoparametric-resonance/swing-spring-pendulum-symbolic.ipynb
vr050714/nonlinear-vibration-seminar
663584d46708857383b637610e54fafa753250e2
CC0-1.0
2021-05-26T05:38:38.000Z
2021-05-26T05:38:38.000Z
autoparametric-resonance/swing-spring-pendulum-symbolic.ipynb
vr050714/nonlinear-vibration-seminar
663584d46708857383b637610e54fafa753250e2
CC0-1.0
autoparametric-resonance/swing-spring-pendulum-symbolic.ipynb
vr050714/nonlinear-vibration-seminar
663584d46708857383b637610e54fafa753250e2
CC0-1.0
Qwen/Qwen-72B
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