I was doing a physics experiment, and I encountered a question which I couldn't answer. The experiment was about using a radar technique to measure the speed of sound. The apparatus was a plastic tube 90cm long and the sound source was a speaker connected to a function generator. The generator was sending a square wave with a low frequency (a couple of Hz) so we were creating sound pulses and we were observing their echoes through a microphone which was connected to an oscilloscope. When the other end of the tube was closed we could clearly see the reflected waves on the oscilloscope screen. However even with an open tube we could see that the waves reflected off the open end. (and they returned with an opposite pressure) Why did this happen?
[Physics] Reflection of sound waves
acousticsboundary conditionsreflectionwaves
Related Solutions
It isn't possible to create an audio source in mid-air using the method you've described. This is because the two ultrasonic waves would create an audible source if the listener were standing at that spot, but those waves would continue to propagate in the same direction afterwards. You would need, as I point out below, some sort of medium which scattered the waves in all directions to make it seem as if the sound were coming from the point at which you interfered the two waves.
It is possible, however, to make the user percieve the sound as coming from a specific location, but it isn't as easy as the author makes it seem. I can think of two different ways. First of all, as described by @reirab, you can get audio frequencies by interfering two sound waves of high frequency. When they interfere they will generate a beat note which has the frequency of the difference between the two frequencies. I.E. if you send a sound beam with frequency $f_1=200\ \text{kHz}$ and another beam with $f_2=210\ \text{kHz}$, the frequency heard in the region where they combine will be $\Delta f-=f_2-f_1=10\ \text{kHz}$ which is in the audio band of humans.
There is an additional difficulty. You will need the sound to come out in a well-defined, narrow (collimated) beam, and this is not terribly easy to do. A typical speaker emits sound in all directions. There are many techniques for generating such beams, but one is to use a phased array.
How can you use this to make a person perceive the sound as coming from a specific point?
Sending Two Different Volumes to the Two Ears
What does it mean to perceive sound as coming from a specific location? Our ears are just microphones with cones which accept sound mostly from one direction (excepting low frequencies). A large part of the way we determine where the sound came from is just the relative volume in our two ears. So, you could use the interference effect described above with beams which are narrow enough that you can target each ear. By using two separate sets of beams targeting each ear with different volumes, you could make the person perceive the sound as coming from a specific location; at least as well as a 3D movie makes a person perceive images in 3D.
Hitting a Material Which Scattered the Sound Isotropically
The second method is to use the same interference effect, but this time combining the two beams at a point where a material scattered the sound waves in all directions. I'm going to be honest, I'm not sure how realistic such materials are, but lets assume they exist for now. If you did so, the two sound beams would be scattered with equal amplitude in all directions and the person you are trying to fool would percieve the sound as coming from this point. This method has the advantage of truly sounding to the person as if the sound came from that direction in all respects including reflections, phasing, etc.
In summary, the idea is definitely possible (maybe there are more ways than I've given), but it isn't as simple as the passage in the book makes it out to be.
Best Answer
See for example: http://www.acs.psu.edu/drussell/Demos/reflect/reflect.html
Reflection from hard vs. soft boundary. Closed tube is like a hard boundary, open tube a soft boundary. In either case there is an impedance discontinuity and hence reflection, however the phase of the reflected wave differs depending on the nature of the discontinuity.