I have an exam question, where they ask me to find the mathematical description of a standing soundwave. They want me to find the displacement wave function $y(x,t)$ and the pressure wave function $p(x,t)$ of a soundwave being reflected at an open end and another being reflected at the closed end of a tube. I know I just have to find the superposition of two waves traveling in the opposite direction.

But what happens with the phase shift for $y(x,t)$ and $p(x,t)$?

I can't find any information in my text book. When a string hits a fixed end/reflected at a node, it experiences a phase shift, because of the force exerted by the wall. But what happens with sound?


2 Answers 2


What you already know about strings is not specific to the behaviour of strings under (co)sinusoidal perturbation. This has to do with the boundary conditions imposed in order to reach a solution to the differential equation (the wave equation in this case).

So, the boundary conditions imposed for the two cases of open end and close end for both displacement and pressure are:

1. Preliminaries

The form of the general one-dimensional wave equation due to D' Alembert is

$$ y \left(x, t \right) = y_{1} \left( c t - x \right) + y_{2} \left( c t + x \right) \tag{1} \label{1} $$

where $y_{1}$ and $y_{2}$ are arbitrary functions with continuous derivatives up to the second order and $c$ is the speed of sound. As long as the relation of the free variables in the argument of the functions have the specified form the two functions constitute solutions to the wave equation.

2. Displacement

  1. Close end: The displacement $y \left( x, t \right) = 0, ~ \forall ~ t$. Since the particles at the closed end cannot move (rigid termination).

We can consider that a positive and a negative travelling wave are superposed. Since we are talking about the same "signal" we can replace $y_{2} \left( x, t \right)$ with $y_{1} \left( x, t \right)$. For the solution to be identically zero for all time we can just negate $y_{2}$ which results to $y_{2} \left( x, t \right) = -y_{1} \left( x, t \right)$. The final form of the solution here is

$$ y \left( x, t \right) = y_{1} \left( c t - x \right) - y_{1} \left( c t + x \right) = 0, ~~~ \forall ~ t \tag{2} \label{2} $$

The negative sign of the "reflected" (or travelling in the opposite direction) wave denotes a phase shift of $\pi$ radians, or $180^{o}$.

  1. Open end: At the open end, the particles are free to move.

Following the same approach we used for the close end, the general form of the solution is

$$ y \left( x, t \right) = y_{1} \left( c t - x \right) + y_{1} \left( c t + x \right) \tag{3} \label{3} $$

What does equation \ref{3} tell us about the displacement though? Since there are no restrictions to the movement of the particles, they are "allowed" to attain the maximum possible amplitude allowed by the superposition of the two solutions ($y_{1} \left( c t - x \right)$ and $y_{1} \left( c t + x \right)$). Thus, the displacement exhibits a maximum at the open end (this makes sense since particles are not constrained in any way).

Following the same mindset as before there seems to be no phase shift for the reflected wave here, which is the case.

3. Pressure

Before we go on to tackle the issue of the pressure at the two ends let us consider the fact that, for a plane travelling wave, the pressure is the first spatial derivative of the displacement (this is true for the wave equation and is not specific to air - applies to strings too).

  1. Close end: As already stated, at the close end the particles are immobile. Since the neighbouring particles are free to move they will eventually concentrate "on" the boundary. Concentration of particles equals increase in pressure. As you can understand, there will be a pressure maximum.

From a mathematical point-of-view, considering a (co)sinusoidal solution for the displacement, its derivative will have co-sinusoidal (or sinusoidal if cosine is considered initially) solution. These two functions have the feature that when one exhibits a maximum the other exhibits a zero. Thus, when the displacement is $0$, the pressure is maximum (which as described above is quite meaningful).

  1. Open end: In the same way we described above the pressure maximum at the close end boundary, the pressure is $0$ at the open end.

Considering again the derivative of the displacement, one can reach the same result (zero pressure) for the open end. This, again, makes sense since there is nothing to constrain the particle displacement and allow pressure build-up at the open end.

Regarding the phase shift for the pressure, one has to consider that the solution depicted in equation \ref{1} is valid for the wave equation irrespective of the medium and for all quantities. Thus, this applies to the pressure in air too. Following the same approach we did for the displacement, one can reach the conclusion that, in order for the value of the pressure (solution) to be identically $0$ for all time $t$, a phase shift of $\pi$ (or $180^{o}$) has to occur at the interface.


I think this: https://en.wikipedia.org/wiki/Reflection_phase_change gives you some of the information you are looking for, i.e. a sound wave propagating through air in a cavity reflects with no phase change at a solid interface, and with a $\pi$-phase change at the open end of the cavity.

What the article doesn't tell you is whether it is talking about pressure or displacement, so here is how I remember it intuitively:

  • The inside of the open end of the cavity has to have the same pressure as outside. Therefore, you get a $\pi$ phase shift for the pressure (so that it cancels) at the open end. However, the displacement has no such limitation.

  • The closed end of the cavity has a fixed position. Therefore, the displacement has to be 0, which means that it experiences a $\pi$ phase shift at the closed end. However, the pressure has no such limitation.

Also, if I remember well, when you solve the wave equations for sound, you'll see that the nodes (i.e. positions with 0 amplitude) for pressure are the positions where you have maximum amplitude for displacement and vice-versa.

Therefore, the conclusion is this:

  • At an open end, pressure is 0 (or more precisely, the difference of pressure with steady state) but displacement is maximum
  • At a closed end, displacement is 0, but pressure is maximum.

Hope it answers all your questions!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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