- #106
- 24,492
- 15,012
I'm really puzzled, why there's still so much confusion. This example is now utmost simple. You have in this case the matrix ##(k_{ij})=\mathrm{diag}(m \omega_1^2,m \omega_2^2)## already diagonalized. Thus the EoM reads
$$\mathrm{d}_t^2 \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=-\begin{pmatrix} \omega_1^2 x_1 \\ omega_2^2 \vec{x}_2 \end{pmatrix}.$$
These are simply two-uncoupled harmonic oscillators.
Now you look for single-frequency solutions with the ansatz
$$\vec{x}(t)=\vec{A} \exp(-\mathrm{i} \omega t).$$
Plugging this into the EoM gives after cancelling the common factor ##\exp(-\mathrm{i} \omega t)##
$$-\omega^2 \vec{A}=-\begin{pmatrix}\omega_1^2 A_1 \\ \omega_2^2 A_2 \end{pmatrix}.$$
Now if ##\omega_1 \neq \omega_2##, you necessarily must have either
$$\omega=\omega_1, \quad A_2=0$$
or
$$\omega=\omega_2, \quad A_1=0.$$
This is not surprising since you simply have the two eigenvectors ##(A_1,0)## and ##(0,A_2)## for the normal modes of the two oscillators. Writing it in terms of the two normalized eigenvectors ##\vec{e}_1=(1,0)## and ##\vec{e}_2=(0,1)## the general solution thus is
$$\vec{x}(t)=\vec{e}_1 (C_{11} \exp(-\mathrm{i} \omega_1 t) +C_{12} \exp(\mathrm{i} \omega_1 t) +\vec{e}_2 (C_{21} \exp(-\mathrm{i} \omega_1 t) +C_{22} \exp(\mathrm{i} \omega_1 t),$$
as you can of course read off directly from the EoM, because the matrix ##\hat{k}## was already given in diagonalized form.
$$\mathrm{d}_t^2 \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=-\begin{pmatrix} \omega_1^2 x_1 \\ omega_2^2 \vec{x}_2 \end{pmatrix}.$$
These are simply two-uncoupled harmonic oscillators.
Now you look for single-frequency solutions with the ansatz
$$\vec{x}(t)=\vec{A} \exp(-\mathrm{i} \omega t).$$
Plugging this into the EoM gives after cancelling the common factor ##\exp(-\mathrm{i} \omega t)##
$$-\omega^2 \vec{A}=-\begin{pmatrix}\omega_1^2 A_1 \\ \omega_2^2 A_2 \end{pmatrix}.$$
Now if ##\omega_1 \neq \omega_2##, you necessarily must have either
$$\omega=\omega_1, \quad A_2=0$$
or
$$\omega=\omega_2, \quad A_1=0.$$
This is not surprising since you simply have the two eigenvectors ##(A_1,0)## and ##(0,A_2)## for the normal modes of the two oscillators. Writing it in terms of the two normalized eigenvectors ##\vec{e}_1=(1,0)## and ##\vec{e}_2=(0,1)## the general solution thus is
$$\vec{x}(t)=\vec{e}_1 (C_{11} \exp(-\mathrm{i} \omega_1 t) +C_{12} \exp(\mathrm{i} \omega_1 t) +\vec{e}_2 (C_{21} \exp(-\mathrm{i} \omega_1 t) +C_{22} \exp(\mathrm{i} \omega_1 t),$$
as you can of course read off directly from the EoM, because the matrix ##\hat{k}## was already given in diagonalized form.