Why antinodes must be present at the free ends of an open organ? (Neglecting end correction)

Question

Why cant a node or less than an antinode be present at the free ends of an open organ pipe. What's the logical reason behind an antinode existing at the open organ pipe? When I asked my teacher this question he answered it by another question "because air particles oscillates with maximum amplitude at the free ends". My question is Why do the air particles oscillate at the free ends with maximum amplitude? Is it necessary that air particles oscillate with max amplitude at the ends?

Answer 1

Personally, I find it easier to think about sound waves as pressure (or density) fluctuations rather than fluctuations in particle displacement. The gauge pressure has to be zero at open ends of the pipe because the pressure is just atmospheric pressure (gauge pressure zero). If you want to convert back to the displacement picture, the pressure and particle displacement are perfectly out of phase with each other, so pressure nodes will be displacement anti-nodes, and vice versa.

Answer 2

1D acoustic waves arises from the isentropic linearized Euler equations, namely

$\left\{\begin{matrix}\partial_t \rho + \rho_0 \partial_x u = 0 \\ \rho_0 \partial_t u - \partial_x P = 0 \end{matrix} \right.$

being $\rho$, $u$, $P$, the variation of density, velocity and pressure around the base state of the fluid at rest $\rho_0$, $u_0 = 0$, $P_0$.

If the process is isentropic, we can evaluate the derivative of pressure $P(\rho,s)$ as

$dP = \underbrace{\left( \dfrac{\partial P}{\partial \rho} \right)_s}_{= c_0^2} d\rho+ \left( \dfrac{\partial P}{\partial s} \right)_{\rho} \underbrace{ds}_{=0} = c_0^2 d \rho$,

where $c_0$ is the speed of sound in the reference base state. Using the last relation, we can write Euler equations as

$\left\{\begin{matrix} \frac{1}{c_0^2}\partial_t P + \rho_0 \partial_x u = 0 \\ \rho_0 \partial_t u - \partial_x P = 0 \end{matrix} \right.$

Now, at a free end in $x=L$:

• pressure is equal to the pressure outside the pipe, and thus its variation is identically

  $P(x=L,t) = 0$;

• if $P(L,t) = 0$ for every time $t$, its partial derivative w.r.t. time is identically equal to zero as well,

  $\partial_t P(L,t) = 0$;

• using the first equation in the Euler system we get

  $0 = \partial_tP(L,t) = - \rho_0 c_0^2 \partial_x u(L,t)$,

  and thus the space derivative of the velocity field is equal to zero. If the solution of the wave equation is a standing wave, this condition coincides with:

  • an "anti-node" for the velocity, i.e. maximum amplitude of oscillations;
  • a node for the pressure field, since $P(L,t) = 0$