Vibrations and Waves

Finding Normal Modes and Normal Co-ordinates

-Finding Normal Modes

Let’s consider the following coupled system:

It can be shown that the coupled equations of motion are as follows, where \omega_{0}= \sqrt{\frac{k}{m}}:

\ddot{x_{1}} +\omega_{0}^{2}(2x_{1}-x_{2})=0

\ddot{x_{2}} +\omega_{0}^{2}(2x_{2}-x_{1})=0

Let’s assume solutions in the form,  (note that we will take only the real part for the final solution):

x_{1} = A_{1}e^{i\omega t}

x_{2} = A_{2}e^{i\omega t}

In vector form we have the following:

\begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix} = \begin{pmatrix} A_{1} \\ A_{2} \end{pmatrix}e^{i\omega t}

Finding the appropriate derivatives (\ddot{x_{1}}, \ddot{x_{2}}), and substituting back into the equations of motions (cancelling the common e^{i\omega t} term) gives:

\begin{pmatrix} -\omega^{2}A_{1} \\ -\omega^{2}A_{2} \end{pmatrix} + \begin{pmatrix} 2\omega_{0}^{2}A_{1}-\omega_{0}^{2}A_{2} \\ -\omega_{0}^{2}A_{1}+2\omega_{0}^{2}A_{2} \end{pmatrix} =0

Adding these vectors, and re-writing in matrix form gives the following:

 \begin{pmatrix} 2\omega_{0}^{2}-\omega^{2} & -\omega_{0}^{2}\\ -\omega_{0}^{2} & 2\omega_{0}^{2}-\omega^{2} \end{pmatrix} \begin{pmatrix} A_{1} \\ A_{2} \end{pmatrix} =0

This now show us (if we multiply both sides of the equation by the inverse of the matrix) a solution where (A_{1}, A_{2})=(0,0), this is however a trivial solution where the masses don’t move, they just stay put. We aren’t interested in this solution, so the only way this doesn’t occur, (A_{1}, A_{2}) \neq (0,0), is if the matrix is singular (has no inverse). This happens when the determinant of the matrix is equal to 0, so we can say that we have solutions when:

\begin{vmatrix} 2\omega_{0}^{2}-\omega^{2} & -\omega_{0}^{2}\\ -\omega_{0}^{2} & 2\omega_{0}^{2}-\omega^{2} \end{vmatrix}=0

\therefore (2\omega_{0}^{2}-\omega^{2})(2\omega_{0}^{2}-\omega^{2})-\omega_{0}^{4}=0

We can now solve for \omega:

\omega^{2}= \omega_{0}^{2}(2\pm 1)

This gives us 4 values for \omega, however because our solution is in the form Ae^{i\omega t} this corresponds to two sets of complex conjugate pairs, so when these solutions are linearly combined they will just give us some other arbitrary coefficient on the amplitude twice the value that you just get considering one value from the pair (real parts of complex conjugates are equal, and we only take the real part). This means we can neglect the negative \omega from each pair of solutions in the following steps. Lets call our solutions \omega_{1} , \omega_{2}:

\omega_{1} = \omega_{0}

\omega_{2} = \sqrt{3}\omega_{0}

These frequencies are Eigenvalues, and correspond to the characteristic frequencies of the normal modes of the system. Substituting the values of \omega back in to the matrix lets us find the corresponding Eigenvectors,

For \omega=\omega_{1}:

 \begin{pmatrix} \omega_{0}^{2}& -\omega_{0}^{2}\\ -\omega_{0}^{2} & \omega_{0}^{2} \end{pmatrix} \begin{pmatrix} A_{1}\\ A_{2} \end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}

Which gives us the Eigenvector \begin{pmatrix}1\\1\end{pmatrix}.

A similar method applied to the Eigenvalue \omega=\omega_{2} gives us the corresponding Eigenvector \begin{pmatrix}1\\-1\end{pmatrix}.

We know that normal modes undergo S.H.M, I can therefore write the normal modes of the system in the following way, where a_{1}, a_{2},\phi_{1},\phi_{2} are defined by initial conditions:

\overrightarrow{\eta_{1}}= a_{1}\begin{pmatrix}1\\ 1\end{pmatrix}cos(\omega_{1}t+ \phi_{1})

\overrightarrow{\eta_{2}}= a_{2}\begin{pmatrix}1\\ -1\end{pmatrix}cos(\omega_{2}t+ \phi_{2})

Remember that initially we assumed a solution to the equations of motion for x_{1} and x_{2} so we say that the general solution is:

\begin{pmatrix}x_{1}(t)\\ x_{2}(t)\end{pmatrix}=a_{1}\begin{pmatrix}1\\ 1\end{pmatrix}e^{i\omega_{1}t}+a_{2}\begin{pmatrix}1\\ -1\end{pmatrix}e^{i\omega_{2}t}

Taking the real part of this (remember we said we would only consider the real part):

x_{1}= a_{1}cos(\omega_{1}t + \phi_{1}) +a_{2}cos(\omega_{2}t + \phi_{2})

x_{2}= a_{1}cos(\omega_{1}t + \phi_{1}) -a_{2}cos(\omega_{2}t + \phi_{2})

This shows us that the complex motion of coupled oscillators is a linear combination (a superposition) of two harmonic normal modes. If only one mode is active (e.g a_{1}=0 or a_{2}=0) the oscillators are either in phase or out of phase:

Let a_{2}=0:

x_{1}= a_{1}cos(\omega_{1}t + \phi_{1})

x_{2}= a_{1}cos(\omega_{1}t + \phi_{1})

The two oscillators move in phase with eachother, however for a_{1}=0:

x_{1}= a_{2}cos(\omega_{2}t + \phi_{2})

x_{2}= -a_{2}cos(\omega_{2}t + \phi_{2})=a_{2}cos(\omega_{2}t + \phi_{2} +\pi)

The phase difference is \pi thus they are out of phase.