Project 10: Coupled Systems

Abstract

The project idea is to work with coupled systems and in my project, I was investigating eigenvalues and eigenvectors in a coupled mass system. The implementation is to analyze three masses connected by springs of equal spring constant. The physical condition was the oscillation of three equal masses coupled by four springs of equal spring constants. We used the python {eig} function to compute the eigenvalues and eigenvectors for this system. Three eigenvalues were found for this system because this sytem involves three coupled masses.

The eigenvalues depended on

k = 10 $Nm^{-1}$ and
m = 1kg.

The eigenvalues of this system were computed to be:$-\omega^2$ = $-34.142, -20.0, -5.858$

Description

First of, we need to understand what and eigenvector and eigenvalues are:

An eigenvector otherwise known as characteristic vector of a linear transformation is a nonzero vector that changes at most by a scalar factor when that linear transformation is applied to it.
An eigenvalue is the factor by which the eigenvector is scaled. They are a special set of scalars associated with a linear system of equations that are sometimes also known as characteristic roots, characteristic values, proper values, or latent roots.

Example

Let A be a linear transformation represented by a matrix A. If there is a vector $x\epsilon$ $R^n \neq0$ such that:

$Ax = \lambda x$, for some scalar lambda, then lambda is the eigenvalue of A with corresponding eigenvector x.

Letting A be a k X k (k by k) square matrix

$$\begin{bmatrix} a{11} & a{12} & \cdots & a{1k} \ a{21} & a{22} & \cdots & a{2k} \ \vdots & \vdots & \ddots & \vdots \ a{k1} & a{k2} & \cdots & a_{kk} \end{bmatrix} $$

with eigenvalue $\lambda$, then the corresponding eigenvectors satisfy $$\begin{bmatrix} a{11} & a{12} & \cdots & a{1k} \ a{21} & a{22} & \cdots & a{2k} \ \vdots & \vdots & \ddots & \vdots \ a{k1} & a{k2} & \cdots & a_{kk} \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ \vdots \ x_k\end{bmatrix} = \lambda \begin{bmatrix} x_1 \ x_2 \ \vdots \ x_k\end{bmatrix} $$

which is equivalent to the homogeneous system $$\begin{bmatrix} a{11}- \lambda & a{12} & \cdots & a{1k} \ a{21} & a{22} - \lambda & \cdots & a{2k} \ \vdots & \vdots & \ddots & \vdots \ a{k1} & a{k2} & \cdots & a_{kk}- \lambda \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ \vdots \ x_k\end{bmatrix} = \begin{bmatrix} 0 \ 0 \ \vdots \ 0\end{bmatrix} $$

The equation above can be written compactly as

$ (A-\lambda I)x=0 $
where I is the identity matrix.

$\det(A-\lambda I)=0$

This equation is known as the characteristic equation of A, and the left-hand side is known as the characteristic polynomial.

Algorithm and Discussion

See a diagram of the system in the figure below

Hooke's law implementation

Using Hooke's Law we can find the force acting on each mass. We compare this to Newton's definition of force. We put all these in matrix forms and we then find out that the eigenvalue is $-\omega^2$.

Method

We using eig function to find the eigenvalues and use the RK4 approach to validate that the eigenvalues computed are correct.

$$k_1 = k_2 = k_3 = k_4 = k$$

$$m_1 = m_2 = m_3 = m$$

$$F_1 = -k\,x_1 +k\,(x_2-x_1)$$

$$F_2 =-k\,(x_2-x_1)+k\,(x_3-x_2)$$

$$F_3 =-k\,(x_3-x_2)+k\,(-x_3)$$

$$\hat{a}\left|t\right\rangle = -\omega^2\begin{bmatrix} x_1 \ x_2 \ x_3\end{bmatrix}e^{-i\omega t}$$

$$\hat{F}\left|x\right\rangle = \begin{bmatrix} -2k & k & 0 \ k & -2k & k \ 0 & k & -2k \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3\end{bmatrix}e^{-i\omega t}$$

$$-m\omega^2\begin{bmatrix} x_1 \ x_2 \ x_3\end{bmatrix}e^{-i\omega t} = \begin{bmatrix} -2k & k & 0 \ k & -2k & k \ 0 & k & -2k \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3\end{bmatrix}e^{-i\omega t}$$

$$\begin{bmatrix} \frac{2k}{m} & \frac{-k}{m} & 0 \ \frac{-k}{m} & \frac{2k}{m} & \frac{-k}{m} \ 0 & \frac{-k}{m} & \frac{2k}{m} \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3\end{bmatrix} = \omega^2\begin{bmatrix} x_1 \ x_2 \ x_3\end{bmatrix}$$

$$\begin{bmatrix} \frac{2k}{m} - \omega^2 & \frac{-k}{m} & 0 \ \frac{-k}{m} & \frac{2k}{m} - \omega^2 & \frac{-k}{m} \ 0 & \frac{-k}{m} & \frac{2k}{m} - \omega^2 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3\end{bmatrix} = 0$$

$$\begin{vmatrix} \frac{2k}{m} - \omega^2 & \frac{-k}{m} & 0 \ \frac{-k}{m} & \frac{2k}{m} - \omega^2 & \frac{-k}{m} \ 0 & \frac{-k}{m} & \frac{2k}{m} - \omega^2 \end{vmatrix} = 0 $$

Method Implementation and Code

Method: We use the eig function to find eigenvalues and eigenvectors then we use to Rk4 to validate the values we computed. We also investigate the motion of each masses with subplots.

%matplotlib inline import pandas as pd import matplotlib.pyplot as pl import numpy as np import sympy as sp from numpy import linalg as LA k = 10 m = 1.0 Mat = np.array([[-(2*k)/m, k/m, 0],[k/m, -(2*k)/m, k/m],[0, k/m, -(2*k)/m]]) vals, vecs = LA.eig(Mat) print("eigenvalues:") print(np.around(vals,3)) print("eigenvector:") print(np.around(vecs, 3))

With the eigenvalues and eigenvector that we just got using the "eig" function, we can validate using the Rk4.

L = 10 k = 10 m = 1.0 dt = 0.03 t = 0.0 x1i =0.5*L/10 # initial displacement from equil, m1 v1i = 0.0 # initial vel, m1 x2i = 0.71*L/10 # initial displacement from equil, m2 v2i = 0.0 # initial vel, m2 x3i = 0.5*L/10 # initial displacement from equil, m3 v3i = 0.0 # initial vel, m3 s = np.array([x1i, v1i, x2i, v2i, x3i, v3i]) # initial state def derivs_3m(s, t): x1=s[0] # get the variables from the state v1=s[1] x2=s[2] v2=s[3] x3=s[4] v3=s[5] a1 = (-(2*k*x1) + k*x2)/m a2 = (-(2*k)*x2 + k*x1 + k*x3)/m a3 = (k*x2 - (2*k)*x3)/m return np.array([v1, a1, v2, a2, v3, a3]) def RK4Step(s, dt, t, derivs): """ Taking a single RK4 step. """ f1 = derivs(s, t) f2 = derivs(s+f1*dt/2.0, t+dt/2.0) f3 = derivs(s+f2*dt/2.0, t+dt/2.0) f4 = derivs(s+f3*dt, t+dt) return s + (f1+2*f2+2*f3+f4)*dt/6.0 x1list=[s[0]] x2list=[s[2]] x3list=[s[4]] tlist=[0.0] t=0.0 while t<6: s = RK4Step(s, dt, t, derivs_3m) t += dt x1list.append(s[0]) x2list.append(s[2]) x3list.append(s[4]) tlist.append(t) pl.subplot(221) pl.ylabel("x1") pl.title("Motion of Coupled Masses") pl.plot(tlist,x1list,label="x1") pl.subplot(222) pl.ylabel("x2") pl.xlabel("t") pl.plot(tlist,x2list,label="x2") pl.subplot(223) pl.ylabel("x3") pl.xlabel("t") pl.plot(tlist,x3list,label="x3")

Using curvefit function to find one out of the three of the normal mode frequencies.

from scipy.optimize import curve_fit def cosFit(t, A, omega, phi): """ Function def for a cosine fit """ return A*np.cos(omega*t+phi) x1a=np.array(x1list) ta=np.array(tlist) popt, pcov = curve_fit(cosFit, ta, x1a, p0=(0.5, np.sqrt(5.858), 0.0)) A=popt[0] omega=popt[1] phi=popt[2] print("A =>", A) print("omega**2 =>", omega**2) print("phi =>", phi) pl.title('Fit to find frequency.') pl.xlabel('t') pl.ylabel('x1') pl.plot(ta, cosFit(ta, A, omega, phi), 'b-', label="fit") pl.plot(ta, x1a, 'r.', label='data') pl.legend()

def cosFit(t, A, omega, phi): """ Function def for a cosine fit """ return A*np.cos(omega*t+phi) x2a=np.array(x2list) ta=np.array(tlist) popt, pcov = curve_fit(cosFit, ta, x2a, p0=(0.7071, np.sqrt(5.858), 0.0)) A=popt[0] omega=popt[1] phi=popt[2] print("A =>", A) print("omega**2 =>", omega**2) print("phi =>", phi) pl.title('Fit to find frequency.') pl.xlabel('t') pl.ylabel('x2') pl.plot(ta, cosFit(ta, A, omega, phi), 'b-', label="fit") pl.plot(ta, x2a, 'r.', label='data') pl.legend()

from scipy.optimize import curve_fit def cosFit(t, A, omega, phi): """ Function def for a cosine fit """ return A*np.cos(omega*t+phi) x3a=np.array(x3list) ta=np.array(tlist) popt, pcov = curve_fit(cosFit, ta, x3a, p0=(0.7071, np.sqrt(5.858), 0.0)) A=popt[0] omega=popt[1] phi=popt[2] print("A =>", A) print("omega**2 =>", omega**2) print("phi =>", phi) pl.title('Fit to find frequency.') pl.xlabel('t') pl.ylabel('x3') pl.plot(ta, cosFit(ta, A, omega, phi), 'b-', label="fit") pl.plot(ta, x3a, 'r.', label='data') pl.legend()

We can see that one of the normal mode frequency was the squareroot of one of the eigenvalues.

Conclusion

The implementation of this lab was to analyze three masses connected by springs of equal spring constant. The physical condition was the oscillation of three equal masses coupled by four springs of equal spring constants. We used the python {eig} function to compute the eigenvalues and eigenvectors for this system. Three eigenvalues were found for this system because this sytem involves three coupled masses.

The eigenvalues depended on

k = 10 $Nm^{-1}$ and
m = 1kg.

The eigenvalues of this system were computed to be:$-\omega^2$ = $-34.142, -20.0, -5.858$

The eigenvector of the system was computed to be: $$\begin{bmatrix} 0.5 & 0.707 & 0.5 \ -0.707 & 0 & 0.707 \ 0.5 & -0.707 & 0.5 \end{bmatrix}$$

Also multiplying both $$k$$ and $$m$$ by the same number still gives us the same wigenvalues.

Project 10: Coupled Systems

Abstract

Description

Example

Algorithm and Discussion

Hooke's law implementation

Method

Method Implementation and Code

Conclusion

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