Sound Waves Legacy Problem #28 Guided Solution

Here we have a question about an organ pipe that's acting as a closed-in resonator. And when an organ pipe acts as a closed-in resonator, one of the ends of the long pipe is closed to the surrounding air, forcing air within the pipe to have a displacement node at that position. In other words, air is hardly vibrating at all at that position. The other end of the pipe actually acts as an open end, and so it has one closed end and one open end with a vibrational antinode at that open end. We're given the speed of sound within the air column, 343 meters per second. We're given the length of the organ pipe, 2.29 meters. We're asked to determine a frequency, a specific one, the fifth highest frequency that this pipe can produce. There's a few ways to do this problem, but my approach is going to be to first find the first harmonic frequency and then use it to find the fifth highest frequency. Now to get that first harmonic frequency, I'm going to need to use the wave equation, but first I need to get the wavelength of this first harmonic. So I'm going to need to recognize that within the closed-in air column, there's one quarter of a wavelength within the 2.29 meters length of the air column. So that will get me, I'll be able to use that relationship to find the wavelength, the wave equation to find the frequency, and then my understanding of the various frequencies or harmonic numbers of a closed-in air column to get the fifth highest. So here we go, I'm going to begin by stating 2.29 meters equal one quarter wavelength, and I'm going to solve for wavelength using that equation. When I do, I get 9.16 meters as my wavelength for the first harmonic frequency. Now I'm going to use the wave equation, B equal F lambda, in order to calculate the frequency of this lowest harmonic, first harmonic. So V equal F lambda rearranges to F equal V divided by lambda, where the V is the speed and the lambda is the wavelength. So F1 equal 343 divided by 9.16 meters. That gets me 37.4454 hertz as the frequency of the first harmonic. Now is where some thinking comes in. You have to think, well what harmonics can a closed-in air column produce? And it produces only odd harmonics, no even harmonics. A discussion of that topic can be found by clicking the link back to the physics tutorial where you read about closed-in air columns. So closed-in air columns can produce the first, the third, the fifth, the seventh, the ninth, the eleventh, etc. And of the frequencies it can produce, the fifth highest is actually the ninth harmonic frequency. So I'm going to take this first harmonic frequency of 37.4454 hertz and I'm going to multiply by 9 and what I do I get 337.01 hertz. I can round that to three significant digits and the answer is 337 hertz.