Normal modes in a coupled system

[rahaverhma]

Why would normal modes occur in the coupled oscillator system I.e. why the parts of system would oscillate with constant angular frequency and constant phase difference ?

 

[tech99]

Why would normal modes occur in the coupled oscillator system I.e. why the parts of system would oscillate with constant angular frequency and constant phase difference ?

If the network were to alter the frequency, it would need to change the shape of the wave whilst doing so. For example, a frequency modulated wave is not sinusoidal.
To alter the shape of a wave the network would need to be either non linear, which it is not (it has only linear LCR components), or have varying parameters, such as a varying C.
If the network has a double response, such as with over-coupled resonators, it can, however, start oscillating on one of two frequencies when excited by an impulse.

 

[rahaverhma]

Actually I had come across a question of 2 springs ,2 blocks suspended by wall under gravity.And I think the shape of wave would not change because there is only vertical oscillations .Yeah,the system has linear equation ,but what do we mean by this ?

 

[tech99]

Actually I had come across a question of 2 springs ,2 blocks suspended by wall under gravity.And I think the shape of wave would not change because there is only vertical oscillations .Yeah,the system has linear equation ,but what do we mean by this ?

Displacement is proportional to force (Hooke's Law).

 

[rahaverhma]

Displacement is proportional to force (Hooke's Law).

But how can I know that they both will be oscillating at same frequencies.And which u hv said that only tells about the frequency of parts of a system.I mean omega is unique for the components .

 

[vanhees71]

Well you have a potential like
$$V(\vec{x})=\frac{1}{2} \sum_{jk} x_j x_k D_{jk},$$
where $D_{jk}$ is a positive definite symmetric matrix.

Now you can always diagonalize such a matrix by choosing another Cartesian basis (i.e., by a rotation of the original Cartesian basis). In this basis you have
$$V(\vec{x})=\sum_{j} \frac{D_j}{2} x_j^2.$$
The force components in this frame are
$$F_j=-D_j x_j$$
(with NO Einstein summation implies).

This means in this frame the components are not coupled and thus each oscillates with its fixed angular frequency $\omega_j=\sqrt{D_j/m}$.

 

[Dr.D]

In one dimensional motion (vertical only), two masses implies two degrees of freedom. When you set up the eigen problem, you will get two roots (two natural frequencies) and two eigenvectors. Each eigenvector describes a mode shape. The actual physical motion of the masses in free vibration is a linear combination of those mode shapes, not just one or the other.

 

[nasu]

Why would normal modes occur in the coupled oscillator system I.e. why the parts of system would oscillate with constant angular frequency and constant phase difference ?

The system would not do this unless you set up the proper initial conditions in such a way as to determine the system to oscillate in a normal mode.
If you just excite the system in a random way, it will oscillate in some complex way, with components possibly stopping at some instants and starting again. But no matter how complicated, the motion can be seen as a combination of normal modes.

As an analogy,a vector may be described as a sum of its components along the axes of coordinates. You can pick up a coordinate system so that the vector is all along x-axis but this is just a special case. And so is a system oscillating in a normal mode.

 

[vanhees71]

Of course, the general solution of the homogeneous linear ODEs are the superpositions of the eigenvectors of the corresponding linear differential operator (aka eigenmodes).