waves - Question on open organ pipe

Although open organ pipe is open on both ends, how standing waves are produced in a open organ pipe. Can someone explain with more clarity?

Answer

You may find by starting from first principles, or by consulting external resources that pressure waves in air (in one dimension) are governed by the wave equation

$$\frac{\partial^2 p}{\partial x^2} - \frac{1}{v^2} \frac{\partial^2 p}{\partial t^2} = 0$$

where $x$ is a position and $t$ is the time, and $p$ denotes the pressure difference away from equilibrium. To solve this, let us write the solution as a Fourier transform$^{[a]}$

$$p(x,t) = \int_k \int_\omega \tilde{p}(k,\omega) \cos(kx - \omega t) \frac{d\omega}{2\pi}\frac{dk}{2\pi} \, .$$ Plugging this into the wave equation, we find

$$(-k^2 + \omega^2 / v^2) \tilde{p}(k,\omega) = 0 \, .$$

For each value of $k$ and $\omega$ one of two things must be true: either $\tilde{p}(k,\omega)=0$ or $\omega^2/v^2 - k^2 = 0$. The case where $\tilde{p}(k,\omega)=0$ are trivial: there is no wave. In the other case, for a particular choice of $k$ and $\omega$ we have a wave with the specific form $$p(x, t) = M \cos(kx - \omega t ) = M\cos(k(x \pm vt)) = \cos(kx)\cos(\omega t) \pm \sin(kx)\sin(\omega t)\, .$$ Since the wave equation is linear, these solutions can be summed together to produce other solutions. Furthermore, any solution can be written as a sum of these types of solutions.

Now consider an organ pipe which we stimulate at frequency $\omega$. In this case, the linear combination can be made up of only the following two parts: $$ \begin{align} p(x,t) &= p_{\text{left}}(x,t) + p_{\text{right}}(x,t) \\ \text{where} \qquad p_{\text{left}}(x,t) &= M_{\text{left}} \cos(kx + \omega t) \\ \text{and} \qquad p_{\text{right}}(x,t) &= M_{\text{right}} \cos(kx - \omega t) \\ \text{so} \qquad p(x,t) &= (M_{\text{left}} + M_{\text{right}})\cos(kx)\cos(\omega t) \\ &+ (M_{\text{right}} - M_{\text{left}}) \sin(kx)\sin(\omega t)\, . \end{align} $$

Suppose the pipe extends from $x=-L/2$ to $x=L/2$. The two ends are open to the outside air, so at those points $p \approx 0$. These are our boundary conditions. From the form of $p(x,t)$ you can convince yourself that the only possible solution is when $M_{\text{left}} = M_{\text{right}}$ and $k=\pi/L$.$^{[b]}$ Putting that in gives

$$p(x,t) \propto \cos(\pi x / L) \cos(\omega t)\, .$$

Stop and consider the meaning of this equation for a minute. It is a vibration of air pressure inside the pipe with cosine spatial profile and also cosine oscillation in time. In other words, the pipe, even with its open ends, can made a vibration at the frequency $\omega$. Note that the pressure difference at the ends of the pipe is zero by construction (i.e. our choice of $k$), yet there are still vibration modes!

$[a]$ I dropped a phase here which, if you put it in and follow everything through, doesn't change the argument.

$[b]$ Actually there are more possible values of $k$. They correspond to higher modes of the vibrating air wave. See if you can figure out what I mean :)