# Which arrangement of pipe and sources can create a standing wave?

A loudspeaker emitting sound of frequency f is placed at the open end of a pipe of length l which is closed at the other end. A standing wave is set up in the pipe.

A series of pipes are then set up with either one or two loudspeakers of frequency f. The pairs of loudspeakers vibrate in phase with each other. Which pipe contains a standing wave?

The correct answer is D; however, I think that all choices are correct. I have learnt that a standing wave is a superposition of two waves moving in opposite directions, each having the same amplitude and frequency. So, any closed pipe as shown in the question must create a standing wave because a reflected wave from the closed end has the same frequency and amplitude as a wave created from the source. Also, I'm confused that I can still draw a formation of standing waves inside all pipes which have opened ends on both sides, too. Could you explain the way to approach this problem ,please?

Any help would be much appreciated.

• The first setup gives you the relation between f and l. Commented Jun 5, 2016 at 4:52

Remember that the air at the closed end can't move, so the amplitude at that location is always zero: it is a node for the standing wave. The displacement or amplitude will be maximal at the speaker, it is an antinode. So the length of the pipe determines the wavelength. The distance between node and antinode of a wave is $\lambda/4$, so $l=\lambda/4$. The distance between two antinodes (or between two nodes) is $\lambda/2$ or $2l$. (see this link for more info).

Open ends (and speakers) are antinodes, closed ends are nodes. So to get a standing wave (using the same frequency), the distance between 2 speakers or between a speaker and open end must be $2l$ (or $4l$, $6l$..) and the distance between speaker and closed end must be $l$ (or $3l$, $5l$ ...). Only case D fits those conditions.

Note that I assumed that f is the lowest frequency (the fundamental frequency) that will produce a standing wave. f could also be the first (second, third..) harmonic, in which case $\lambda=4l/3$ ($4l/5$, $4l/7$ ...). But that doesn't matter in this specific case, since the options given only include lengths of $l$ and $2l$, and you know that $l$ is the distance between a node and an antinode. Which also means that at a distance of $2l$ from an antinode will be an antinode.