1
$\begingroup$

Consider two waves $y_1,y_2$ travelling in opposite directions with equations $$y_1(x,t) = A \sin(\omega t - kx) \\ y_2(x,t) = A \sin(\omega t + kx) $$ That create the following standing wave $$y_s(x,t) = y_1 + y_2 = 2A\cos(kx)\sin (\omega t)$$

Consider, now, a point $P$ at $x = x_0$ (without loss of generality let $x_0 > 0$) whose motion as a function of time is described as

$$y_P = y(x_0,t) = \left\{ \begin{array}{lr} A \sin(\omega t + kx_0) & : 0 < t < kx_0 / \omega \\ 2A\cos(kx_0)\sin (\omega t) & : t \ge kx_0 / \omega \end{array} \right. $$

It is easy to show that $y_P$ is everywhere continuous, however is not always differentiable at $t_0 = kx_0 / \omega$.

My question: In case the elastic medium is not massless then what is the velocity of $P$ at time $t= t_0$? What's the physical meaning of non-differentiability of $y_P$? (The medium has infinite length)

$\endgroup$
3
$\begingroup$

The equations you wrote down describe a wave that is traveling one way, then gets reflected and travels back the other way. The leading edge of the returning wave changes the slope of the wave - just before it becomes a full standing wave it is a mixture of the traveling wave (to the right) and the leading edge of the reflected wave.

At that leading edge, there is a discontinuity in the slope of the wave, and thus "infinite acceleration" locally. This is why you are having trouble with the differentiation - the kink in the amplitude implies a singularity.

In a "real" material, such a discontinuity would result in a degree of dispersion - depending on the properties of the material - since the material would not support infinite force (it either tears the material, or the slope becomes more gradual). I don't know that you can define the velocity at the exact point where the discontinuity occurs, since acceleration is infinite.

Just one of those things where mathematics doesn't work because it no longer reflects real world physics.

$\endgroup$
  • 1
    $\begingroup$ My thought was that the medium will "squeeze" at that point in order to tackle this discontinuity in velocity $\endgroup$ – bolzano Mar 6 '15 at 19:12
0
$\begingroup$

I believe it means that there is a kink or cusp in the pattern the wave makes at that point in time, presumably due to the fact that there was some external disturbance needed to make that change happen.

$\endgroup$

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.