Today, we begin with two coupled oscillators. The two oscillators \(m_1\) and \(m_2\) are connected with spring constant \(k_{12}\). Spring constant \(k_1\) connects \(m_1\) to a fixed position and \(k_2\) connects \(m_2\) to a fixed position on the other end.
The first is to figure out Newton's equations governing each mass. We do that by pulling on one of the masses(say \(m_2\)) to the right by distance \(x_2\). Now what effect will this have on \(m_1\)? Well, \(m_1\).will be displaced from its position by \(x_1\) distance. The force on spring constant \(k_1\) will want to return the spring back to equilibrium so it will be \(-k_1x_1\). Spring constant \(k_{12}\) will want to return to equilibrium so it will be \(-k_{12}x_{1}\). But there is a third force( the one causing all the stretching), this acts opposite to the two opposite forces and is equal to \(k_{12}x_2\). All this analysed from the perspective of \(m_1\). Thus far we have this for \(m_1\): \begin{equation} m_1\frac{d^2x_1}{dt^2}= -k_1x_1 - k_{12}x_1 + k_{12} x_2 (1)\end{equation}
The second step is to analyse the system from the perspective of \(m_2\). So perturb the system by pulling \(m_1\) to the left by amount \(x_1\) and watching the springs holding \(m_2\). What do we see? First, we see \(k_2)\) pulling \(m_2\) back to equilibrium so the force is \(-k_2 x_2\) and so does \(k_{12}\) with force \(-k_{12}x_2\). Again there is the last force that is causing all the stretching and is acting in opposition to the first two forces, its amount is \(k_{12}x_1\). Thus we arrive at the last equation for the second mass, namely: \begin{equation} m_2\frac{d^2x_2}{dt^2}= -k_2x_2 - k_{12}x_2 + k_{12} x_1 (2) \end{equation}
If you place the first two equations underneath each other you will recognize matrix multiplication. We can thus make the following matrix equation:
\begin{align}
\begin{pmatrix}m_1 x_1^{''}\\m_2 x_2^{''}\end{pmatrix}= \begin{pmatrix}-(k_1+k_{12})& k_{12}\\ k_{12}&-(k_2+k_{12})\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}
\end{align}.
It is important to notice that the matrix is symmetric. This implies that it has real eigenvalues and its eigenvectors are orthogonal, this will come into play later on. We then the make an ansatz , namely that the solutions are of the form \(x_1\) = \(A_1\)\(e^{i\omega t}\) and \(x_2\)= \(A_2\)\(e^{i\omega t}\) and plug into our matrix equation. The exponetial function will drop out living the following:
\begin{align}
\begin{pmatrix}-m_1 A_1\omega ^2\\-m_2 A_2\omega^2\end{pmatrix}= \begin{pmatrix}-(k_1+k_{12})& k_{12}\\ k_{12}&-(k_2+k_{12})\end{pmatrix}\begin{pmatrix}A_1\\A_2\end{pmatrix}
\end{align}.
The above equation can be re-arranged to give the following :
\begin{align}
0= \begin{pmatrix}(k_1+k_{12}-m_1\omega^2) & k_{12}\\ k_{12}&(k_2+k_{12}-m_2\omega^2)\end{pmatrix}\begin{pmatrix}A_1\\A_2\end{pmatrix}
\end{align}.
These last few steps are usually the clear steps in text books. One can easily follow the math, but I decided to re derive for the sake of clarity and mellifluous flow of logic. Usually textbooks jump to saying a non trivial solution is got by making the determinate of the 2 by 2 matrix equal to zero but why is this? How does the determinate suddenly rear its head?
Consider the following matrix equation Ax=b. If A is a non-singular matrix then this equation has only one solution. Now make b=0. It still has one solution and due to the invertible matrix theorem the only solution is x being the zero vector. But that is the trivial solution to our problem. So instead we look for matrices that are singular and thus have more than just the zero vector as the solution. Now, how do we find these matrices? Simple, their defining characteristic is that their determinant is zero. Thus we set the determinat of our matrix equal to zero.
I shall now skip to the solution. You already know from the hundreds of materials out there that one gets two kinds of solutions. 1) Both oscillators have the same amplitude 2) Out of phase with each other. This means we can write the solution thusly:
\(x_a(t)\)= \(A_1\) \(\cos \omega_1 t\) + \(A_1\) \(\cos \omega_2 t\) and \(x_b(t )\)= \(A_1\) \(\cos \omega_1 t\) - \(A_1\) \(\cos \omega_2 t\). Again put one equation underneath the other and a matrix equation suddenly appears:
\begin{align}
\begin{pmatrix}x_a(t) \\x_b(t)\end{pmatrix}= \begin{pmatrix}1 & 1 \\ 1 & - 1\end{pmatrix}\begin{pmatrix} \cos \omega_1 t \\ \cos \omega_2 t \end{pmatrix}
\end{align}
In this form it becomes apparent that there must be some transformation we can apply so that we are not taking linear combination of cosines, this transformation will help us decouple the two cosines. Why is it apparent? The matrix with all the 1s is diagonalizable.
This is the stage where in texts all logic breaks down and the author brings up some transformation seemingly out of thin but that does the trick. The student is then left wondering what if I had 3 or 4 oscillators how could I figure out the transformation. That is the subject of the next post. What we want to understand is a general and systematic way of getting at these so called "normal co-ordinates".
The first is to figure out Newton's equations governing each mass. We do that by pulling on one of the masses(say \(m_2\)) to the right by distance \(x_2\). Now what effect will this have on \(m_1\)? Well, \(m_1\).will be displaced from its position by \(x_1\) distance. The force on spring constant \(k_1\) will want to return the spring back to equilibrium so it will be \(-k_1x_1\). Spring constant \(k_{12}\) will want to return to equilibrium so it will be \(-k_{12}x_{1}\). But there is a third force( the one causing all the stretching), this acts opposite to the two opposite forces and is equal to \(k_{12}x_2\). All this analysed from the perspective of \(m_1\). Thus far we have this for \(m_1\): \begin{equation} m_1\frac{d^2x_1}{dt^2}= -k_1x_1 - k_{12}x_1 + k_{12} x_2 (1)\end{equation}
The second step is to analyse the system from the perspective of \(m_2\). So perturb the system by pulling \(m_1\) to the left by amount \(x_1\) and watching the springs holding \(m_2\). What do we see? First, we see \(k_2)\) pulling \(m_2\) back to equilibrium so the force is \(-k_2 x_2\) and so does \(k_{12}\) with force \(-k_{12}x_2\). Again there is the last force that is causing all the stretching and is acting in opposition to the first two forces, its amount is \(k_{12}x_1\). Thus we arrive at the last equation for the second mass, namely: \begin{equation} m_2\frac{d^2x_2}{dt^2}= -k_2x_2 - k_{12}x_2 + k_{12} x_1 (2) \end{equation}
If you place the first two equations underneath each other you will recognize matrix multiplication. We can thus make the following matrix equation:
\begin{align}
\begin{pmatrix}m_1 x_1^{''}\\m_2 x_2^{''}\end{pmatrix}= \begin{pmatrix}-(k_1+k_{12})& k_{12}\\ k_{12}&-(k_2+k_{12})\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}
\end{align}.
It is important to notice that the matrix is symmetric. This implies that it has real eigenvalues and its eigenvectors are orthogonal, this will come into play later on. We then the make an ansatz , namely that the solutions are of the form \(x_1\) = \(A_1\)\(e^{i\omega t}\) and \(x_2\)= \(A_2\)\(e^{i\omega t}\) and plug into our matrix equation. The exponetial function will drop out living the following:
\begin{align}
\begin{pmatrix}-m_1 A_1\omega ^2\\-m_2 A_2\omega^2\end{pmatrix}= \begin{pmatrix}-(k_1+k_{12})& k_{12}\\ k_{12}&-(k_2+k_{12})\end{pmatrix}\begin{pmatrix}A_1\\A_2\end{pmatrix}
\end{align}.
The above equation can be re-arranged to give the following :
\begin{align}
0= \begin{pmatrix}(k_1+k_{12}-m_1\omega^2) & k_{12}\\ k_{12}&(k_2+k_{12}-m_2\omega^2)\end{pmatrix}\begin{pmatrix}A_1\\A_2\end{pmatrix}
\end{align}.
These last few steps are usually the clear steps in text books. One can easily follow the math, but I decided to re derive for the sake of clarity and mellifluous flow of logic. Usually textbooks jump to saying a non trivial solution is got by making the determinate of the 2 by 2 matrix equal to zero but why is this? How does the determinate suddenly rear its head?
Consider the following matrix equation Ax=b. If A is a non-singular matrix then this equation has only one solution. Now make b=0. It still has one solution and due to the invertible matrix theorem the only solution is x being the zero vector. But that is the trivial solution to our problem. So instead we look for matrices that are singular and thus have more than just the zero vector as the solution. Now, how do we find these matrices? Simple, their defining characteristic is that their determinant is zero. Thus we set the determinat of our matrix equal to zero.
I shall now skip to the solution. You already know from the hundreds of materials out there that one gets two kinds of solutions. 1) Both oscillators have the same amplitude 2) Out of phase with each other. This means we can write the solution thusly:
\(x_a(t)\)= \(A_1\) \(\cos \omega_1 t\) + \(A_1\) \(\cos \omega_2 t\) and \(x_b(t )\)= \(A_1\) \(\cos \omega_1 t\) - \(A_1\) \(\cos \omega_2 t\). Again put one equation underneath the other and a matrix equation suddenly appears:
\begin{align}
\begin{pmatrix}x_a(t) \\x_b(t)\end{pmatrix}= \begin{pmatrix}1 & 1 \\ 1 & - 1\end{pmatrix}\begin{pmatrix} \cos \omega_1 t \\ \cos \omega_2 t \end{pmatrix}
\end{align}
In this form it becomes apparent that there must be some transformation we can apply so that we are not taking linear combination of cosines, this transformation will help us decouple the two cosines. Why is it apparent? The matrix with all the 1s is diagonalizable.
This is the stage where in texts all logic breaks down and the author brings up some transformation seemingly out of thin but that does the trick. The student is then left wondering what if I had 3 or 4 oscillators how could I figure out the transformation. That is the subject of the next post. What we want to understand is a general and systematic way of getting at these so called "normal co-ordinates".
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