Let's consider $n$ cuboids moving without friction, each of mass $m_i$. Each wo neighboring cuboids are connected with a spring of the coefficient $k$.

|----|      |---|    |----|     |----|    |-----|
|    |\/\/\/|   |\/\/|    | ... |    |/\/\|     |
|----|      |---|    |----|     |----|    |-----|

I came across a assumption that the movement of $i$-th cuboid will be $$x_i = A_i \cos(\omega t + \phi_i)$$ assuming the same $\omega$ for each of the cuboids.

Why can we make such an assumption?

  • 1
    $\begingroup$ I don't know if I'm going to have time to write a good answer about this, but the fundamental reason is that the system has normal modes, which, because the differential equation of this system is time translational invariant, have sinusoidal time dependence. I remember being confused about this for years as a student. Keep asking until you get a complete understanding of this, as it's one of the most important topics in physics. $\endgroup$ – DanielSank Jan 15 '16 at 17:20

It's not that the final solution looks like that. Rather, you are looking for all the solutions of that form (normal modes) for two reasons:

  • They are easy to find

  • You can afterwards decompose any motion into a sum of normal modes. This comes from writing your equations of motion in the normal basis*.

So first you work out the normal modes by assuming a solution like the one you wrote there. Then you write a system of equations that says the coordinates are a linear combination of the normal modes. By evaluating this set of equations at some instant in time, and equating it to a set of initial conditions, you can solve for the amplitude of each normal mode. Here's a complete worked out example.

It is often enough to know what the normal frequencies are, in order to predict attributes like the absorption spectrum of a molecule.

* Your equation has this form: $\frac{d^2\vec x}{dt^2}=A\vec x$. By diagonalizing the symmetric matrix A, you can write it as a diagonal matrix $D$ in a different basis: $A=CDC^T$ ($C$ is the matrix that takes coordinates in the new basis and gives you positions in the old one). Multiplying the original equation by $C^T$ on the left and using $C^TC=\mathbb I$, you get $\frac{d^2}{dt^2}C^T\vec x=DC^T\vec x$. Calling $\vec y=C^T\vec x$ the normal coordinates, you now have a set of uncoupled differential equations: $\frac{d^2y_n}{dt^2}=D_ny_n$. Their solution is $y_n=A_n\sin(\omega_nt+\phi_n)$, with $\omega_n=\sqrt{-D_n}$.

While this proves that you have a basis of normal modes, you don't go through this whole procedure each time. You usually just propose normal mode solutions.

  • 1
    $\begingroup$ It would probably help to show why the normal modes are guaranteed to have the sinusoidal dependence OP mentions. This deserves an in-depth and complete description. $\endgroup$ – DanielSank Jan 15 '16 at 17:24
  • 2
    $\begingroup$ Not all matrices can be diagonalized. How do we know that the matrix $A$ can be diagonalized? $\endgroup$ – marmistrz Jan 16 '16 at 15:58
  • 1
    $\begingroup$ Moreover, a finite mass-spring system will yield a tridiagonal matrix with equal coefficients on each diagonal. There is an analytical solution to this, which has sin function on the eigenvectors as well math.upenn.edu/~kazdan/AMCS602/tridiag-short.pdf The eigenmodes show behavior in very similar manner as sound wave would resonate in a pipe, depending on the boundary conditions. Here the end springs connect to fixed position, whereas in the question, the boundaries were open. $\endgroup$ – Mikael Kuisma Jan 16 '16 at 17:21
  • $\begingroup$ I'll learn about eigenvalues somewhere at the beginning of March (so says the syllabus). Is it possible to show the diagonalization stuff without using the eigenvalues, or should I just simply wait until I learn about them? $\endgroup$ – marmistrz Jan 16 '16 at 20:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.