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For a standing wave (i.e. a wave pattern created from two oppositely-traveling waves with the same amplitude and frequency), stationary nodes are created at certain positions along the wave with uniform spacing between the nodes depending on the length of the system (e.g. the length of a string controls the spacing of the nodes).

A wave can also be the superposition of multiple waves with different frequencies.

My question is:

If we have two transducers, one transducer is producing a wave signal comprised of multiple arbitrary frequencies and the other wave is producing a wave signal comprised of multiple arbitrary frequencies that are not necessarily the same has the other transducer, is there ever a case where we could have non-uniformly spaced nodes?

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I don't think it's possible, so long as the speed of the wave is uniform along the medium and we clearly define what we mean by a standing wave.

Suppose that $\phi(x,t)$ is the displacement of the medium as a function of position and time. The function $\phi$ must satisfy the wave equation: $$ c^2 \frac{\partial^2 \phi}{\partial x^2} = \frac{\partial^2 \phi}{\partial t^2} $$ Now, the characteristic of a standing wave is that all points of the wave oscillate with a common phase, and the positions of the amplitude peaks do not change with time. This suggests that $\phi(x,t)$ should be of the form $$ \phi(x,t) = T(t) X(x) $$ where $T(t)$ is some oscillatory function that tells us the phase of the wave at a given time $t$, and $X(x)$ tells us how big the oscillations are at a given point $x$. (It's possible that there is a less restrictive definition of a standing wave that is possible; this assumption is what does most of the work for what follows.)

Plugging this ansatz into the wave equation leads to the conclusion that $$ c^2\frac1X \frac{d^2 X}{d x^2} = \frac1T \frac{d^2 T}{d t^2}. $$ Since the left-hand side of the equation depends only on $x$ while the right-hand side depends only on $t$, the only way that they can be equal is if they are both constant: $$ c^2\frac1X \frac{d^2 X}{d x^2} = \frac1T \frac{d^2 T}{d t^2} = \alpha. $$ In other words, we must have $$ X'' = \frac{\alpha}{c^2} X \qquad T'' = \alpha T. $$ But if we want $T$ to be oscillatory, then we must have $\alpha < 0$. We can define a new constant $k$ such that $\alpha = - k^2 c^2$, and so $$ X'' = -k^2 X $$ and so we must have $$ X = A \cos (kx + \delta) $$ where $A$ and $\delta$ are determined by the boundary conditions of the system. No matter what $A$ and $\delta$ are, though, the zeroes of $X$ will be evenly spaced with gaps of $\Delta x = \pi/k$ between them.

Note, however, that the above derivation would change if the speed of sound on the string is not uniform. In that case, the above equation would become $$ X'' = -\frac{\alpha}{c^2(x)} X $$ and while the solutions would be oscillatory, they wouldn't have evenly spaced nodes. For example, if the mass density of the string varied linearly with length, then the speed of sound would be different at different points along the string. In this case, the solutions would be Airy functions rather than sinusoids, and the zeros of Airy functions are not evenly spaced.

(You may recognize some of the above argument as part of the standard derivation for separation-of-variables solutions. Effectively, requiring that the solution act like a standing wave implies that the solution to the wave equation is separable.)

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  • $\begingroup$ On the other hand, if you can add two solutions you should get another solution. Does this mean that the addition of two standing waves is a not a standing wave? $\endgroup$
    – user65081
    Aug 20, 2021 at 0:33
  • $\begingroup$ @Wolphramjonny: Yes, that's definitely a loophole. I suppose it depends on how you define a standing wave. Note, however, that most superpositions of two standing waves will not have any nodes at all; you would need all of the constituent modes to have a node in common to get $\phi(x,t) = 0$ for all $t$. I think even in such cases, the nodes would be evenly spaced. $\endgroup$ Aug 20, 2021 at 1:13
  • $\begingroup$ I understand your argument here, but is there ever a case where we could get a node without having a stationary wave? So for instance, could there be a situation where you add to waves together that may be a combination of different frequencies and getting something close to a stationary wave? $\endgroup$
    – CPres
    Aug 27, 2021 at 21:05

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