I have recently started learning about waves. We didn't really formally describe what a wave is, but instead started by looking at a concrete example namely harmonic sinusoidal waves in 1d.

We then introduced the wave equation in 1d and showed that the sinusoidal waves indeed satisfy this equation.

So to my current understanding a 1d wave is a phenomena described by a function which associates to each point in time a scalar displacement at each point on a line, that satisfies the wave equation. So far so good.

Now there is an example where a rod attached to a wall is pulled and we want to analyse the displacement at each point on the rod from its original position over time. Using hooks Law we are able to derive a relationship between the second partial of time and the second partial of space(of the displacement). We then plug this into the wave equation, which gives us an expression for the velocity at which this wave propagates through the rod.

My question is, why are we able to plug these values into the wave equation? What about this example tells us that these displacements will indeed satisfy the wave equation?

  • $\begingroup$ Related: physics.stackexchange.com/q/159021/2451 , physics.stackexchange.com/q/78500/2451 , physics.stackexchange.com/q/201847/2451 and links therein. $\endgroup$ – Qmechanic Jul 17 '18 at 17:00
  • $\begingroup$ Most of those threads were a bit too advanced for me to fully understand. The basic idea I got was that if a system is stable, in the sense that every part of it will try to return to its original state if moved, then a disturbance of this system will result in a wave. Would this characterisation be somewhat true? $\endgroup$ – J.G95 Jul 17 '18 at 20:17
  • $\begingroup$ A more basic starting point for the wave equation might be f(x-vt) which is a waveform moving along the x axis at a speed v. It can be a sine wave or a pulse or any other function that has first and second derivatives. Also the wave equation can be derived from this alone. See physics.stackexchange.com/a/403761/45664 $\endgroup$ – user45664 Jul 18 '18 at 17:03
  • $\begingroup$ Ok, but even here we must still make the assumption that this disturbance spreads through the medium at some constant velocity right? $\endgroup$ – J.G95 Jul 18 '18 at 17:26
  • $\begingroup$ I understand it better know, the problem was that I didn't realise that in any solution function to the differential equation I derived the constants root is indeed the velocity of propagation. $\endgroup$ – J.G95 Jul 18 '18 at 18:46

Your Answer

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

Browse other questions tagged or ask your own question.