According to definition of normal modes, which says if all the different independent parts of a system vibrate at same frequency and their amplitude preserve a fixed ratio then such a motion is a normal mode of that system then since in sinusoidal travelling waves also different parts move with same frequency and different parts preserve a ratio, shouldn't they too be normal modes?
So are sinusoidal traveling waves normal modes?
If the equations of motion of the vibrating system are equivalent to real and symmetric mass and stiffness terms, the normal modes will be real vectors, which means that all parts of the system move in the same phase.
That excludes travelling waves, where there is a phase difference between points in the direction of travel of the wave.
There is a special case, when two (or more) vibration modes have identical frequencies. In that situation, a combination of the different mode shapes with different phases may "look like" a travelling wave. However this may only be a theoretical possibility, because the tolerances in a real-life structures often separate the two theoretically-identical frequencies.
However there are mechanical systems which do have "travelling" normal vibration modes. A simple example is a gyroscope, where the vibration modes include precession and nutation.
In general, the equations of motion of a system rotating with constant angular velocity will include Coriolis terms, if it is modelled in a rotating coordinate system fixed to the undeformed shape of the body. The equations of motion are then Hermitian matrices rather than real symmetric matrices. The eigenvalues (natural frequencies) are still real, but the mode shapes are now complex vectors which can be interpreted as travelling waves.
In general, the speed at which the "mode shape" rotates around the object is different from the rotation speed of the object itself. If the two speeds coincide for some particular rotation speeds, that can have severe consequences for the design of real rotating machinery - for example the so-called "critical speeds" of rotating shafts.
Yes, travelling wave systems have normal modes. In fact, the physics and mathematics of two coupled oscillators is strikingly similar to that of coupled waveguides. I'm most familiar with electrical oscillators, but everything written below applies to any coupled harmonic oscillators.
Consider two electrical oscillators "$a$" and "$b$". Oscillator $a$ has capacitance $C_a$ and inductance $L_a$, and similarly for oscillator $b$. The oscillators are coupled through a capacitance $C_g$ and mutual inductance $L_g$. Each oscillator has a magnetic flux $\Phi$ and an electric charge $Q$.$^{[a]}$
We could study this system using Kirchhoff's laws, but it's a lot easier to convert everything to the Hamiltonian formalism. The Hamiltonian for the system is $$H = \frac{\Phi_a^2}{2 L_a'} + \frac{\Phi_b^2}{2 L_b'} + \frac{Q_a^2}{2 C_a'} + \frac{Q_b^2}{2 C_b'} + \frac{Q_a Q_b}{C_g'} - \frac{\Phi_a \Phi_b}{L_g'} $$ where all those primes on the various constants have to do with the fact that the coupling renormalizes each oscillator's capacitance and inductance. Now we introduce the variables \begin{align} a &= \frac{1}{\sqrt{2}}\left( \frac{\Phi_a}{\sqrt{Z_a'}} + i \sqrt{Z_a'} Q_a \right) \\ b &= \frac{1}{\sqrt{2}}\left( \frac{\Phi_b}{\sqrt{Z_b'}} + i \sqrt{Z_b'} Q_b \right) \end{align} where the impedance $Z$ is defined by $Z \equiv \sqrt{L/C}$. With these variables, the Hamiltonian becomes \begin{align} H &= \omega_a' a^* a + \omega_b' b^* b \\ &-\left( ab + a^* b^* \right) \underbrace{\frac{1}{2} \left( \frac{1}{C_g' \sqrt{Z_a' Z_b'}} + \frac{\sqrt{Z_a' Z_b'}}{L_g'} \right)}_\chi \\ &+\left( a b^* + a^* b \right) \underbrace{\frac{1}{2} \left( \frac{1}{C_g' \sqrt{Z_a' Z_b'}} - \frac{\sqrt{Z_a' Z_b'}}{L_g'} \right)}_g \, . \end{align} Let's remember for a moment what the Hamiltonian means: it provides a way to get the time evolution of the system. In the present case the time dependences come from $$ \dot a(t) = -i \frac{\partial H}{\partial a^*} \qquad \dot b(t) = -i \frac{\partial H}{\partial b^*} \, . $$ Using these equations, we can write a matrix equation for the whole system: $$ \frac{d}{dt} \left( \begin{array}{c} a \\ b \\ a^* \\ b^* \end{array} \right) = -i \left( \begin{array}{cc} \omega_a' & g & 0 & - \chi \\ g & \omega_b' & -\chi & 0 \\ 0 & \chi & -\omega_a' & -g \\ \chi & 0 & -g & -\omega_b' \end{array} \right) \left( \begin{array}{c} a \\ b \\ a^* \\ b^* \end{array} \right) \, . $$ Ok now here's the point: the normal modes and frequencies of the system are precisely the eigenvectors and eigenvalues of that matrix. If the coupling is turned off (i.e. $C_g=0$ and $L_g=0$), then $g = \chi = 0$ and the eigenvalues are $\pm \omega_a'$ and $\pm \omega_b'$, which makes complete sense.
Alright now suppose we have two waveguides "$a$" and "$b$" that are coupled to each other through some mutual capacitance and inductance per length of the waveguide. Denote the rightward and leftward moving amplitudes in waveguide $a$ as $a_\pm$, and similarly for waveguide $b$. If you work it all out, you find that $$ \frac{d}{dx} \left( \begin{array}{c} a_+ \\ b_- \\ a_- \\ b_+ \end{array} \right) = i \left( \begin{array}{cc} k_a' & -g & 0 & \chi \\ g & -k_b' & -\chi & 0 \\ 0 & -\chi & -k_a' & g \\ \chi & 0 & -g & k_b' \end{array} \right) \left( \begin{array}{c} a_+ \\ b_- \\ a_- \\ b_+ \end{array} \right) $$ where the $k$'s are the wave numbers associated with each waveguide (and I should say that the meanings of $g$ and $\chi$ are slightly different than they were for the coupled oscillator case).
Thus the coupled waveguide problem has the same form as the coupled oscillator problem (the signs are different, but that's just because of how we ordered the variables). In both cases we have a system of first-order differential equations, and in both cases the eigenvectors and eigenvalues of the matrix in the differential equation tell us what the normal modes and frequencies (for the oscillator case) or wave numbers (for the waveguide case) of the system are.
$[a]$: There's a direct correspondence between electrical and mechanical oscillators. The electrical flux and charge correspond to position and momentum. Capacitance corresponds to mass, and inductance corresponds to one over the spring constant. Where we use Kirchhoff's laws for the electrical case, we use Newton's law ($F = ma$) in the mechanical case.
