7.8 Longitudinal Standing Waves

We now have two ways of describing longitudinal waves, by the displacement of the particles and by the pressure along the direction of travel.

	$\displaystyle s(x,t)$	$\displaystyle=A\cos(kx-\omega t+\phi_{0})$		(7.144)
	$\displaystyle P(x,t)$	$\displaystyle=P_{\text{max}}\sin(kx-\omega t+\phi_{0}).$		(7.145)

We can see that if we have a longitudinal standing wave, displacement nodes will be pressure antinodes, and pressure nodes will be displacement antinodes.

Let’s look at the allowed wavenumbers for longitudinal waves in a pipe with closed and open ends. In the case with both ends closed, we will have displacement nodes at each end of the pipe (the medium cannot move through the end). Thus the normal modes will have the form

s_{n}(x,t)=A\sin(k_{n}x)\cos(\omega_{n}t+\phi_{0}),

(7.146)

where $k_{n}=\frac{n\pi}{L}$ , $\lambda_{n}=\frac{2L}{n}$ , and $f_{n}=\frac{nv}{2L}=nf_{1}$ . In the case where both ends are open, we have displacement antinodes at each end. This leads to normal modes of the form

s_{n}(x,t)=A\cos(k_{n}x)\cos(\omega_{n}t+\phi_{0}),

(7.147)

with the same allowed wavenumbers. In the case of pipe with one and closed and one end open, there will be a displacement node at the closed end and a displacement antinode at the open end. Depending on which end of the pipe is open, we would use sine or cosine for the spatial part of the normal modes. The allowed wavenumbers are then $k_{n}=\frac{(2n-1)\pi}{2L}$ , $\lambda_{n}=\frac{4L}{2n-1}$ , $f_{n}=\frac{(2n-1)v}{4L}=(2n-1)f_{1}$ . Note that in this case the fundamental mode has a wavelength of $4L$ .