Sound Waves Legacy Problem #24 Guided Solution

Here we have a question about an organ pipe which is acting as a closed-in resonator, meaning that at one of the ends of this long pipe it's closed off to the surrounding air, forcing air inside the pipe to be acting as a displacement node, not vibrating at all, and forcing air at the opposite end to be a displacement anti-node, vibrating vigorously in and out of the end of that open-end tube. The length of this tube is 83 centimeters, and sound waves travel at 350 meters per second. So I know L equals 83 centimeters, and that's L in that wavelength, and I know the speed of waves, 350 meters per second. What I'm asked to calculate is the fundamental frequency of this organ pipe and the frequency of the next two harmonics of this closed-in organ pipe. So for the fundamental frequency, for F1 value, in order to calculate it, I need to use the wave equation, and I need to know the speed, which is given, and the wavelength of the first harmonic, which is not given. So if I picture the first harmonic wave pattern as a node at one end and an anti-node at the opposite end, and it has a quarter of a wavelength within the length of the tube, and if you need help with that, you should click the link back to the standing wave patterns for closed-in air columns at the physics classroom tutorial and take a quick read. So 83 centimeters equal one quarter wavelength. I multiply both sides of that equation by 4, I get the wavelength equal 332 centimeters, or 3.32 meters. That's the wavelength of the first harmonic, and I need to use it to calculate the frequency of the first harmonic. Using the wave equation, V equal F times lambda, I can rearrange it to F equal V divided by lambda, where V is the speed and lambda is the wavelength. Now I can say F of the first harmonic equal 350 divided by the 3.32 meters. I end up with a frequency of 105.4 hertz, and I can round that to two significant digits, such that it becomes 110 hertz. Now for part B of the problem, I need to find the frequencies of the next two highest harmonics. Now for closed-in air columns, it produces what we refer to as odd harmonics. In other words, there's no second harmonic, or fourth, or sixth harmonic. Only odd numbers harmonics. So the second highest frequency is not actually the second harmonic, it's the third harmonic for a closed-in air column. And the third highest frequency is not the third harmonic, it's actually the fifth harmonic. And to find the frequency of any harmonic, I need to take the frequency of the first and multiply it by an integer known as the harmonic number, which for the third harmonic is 3, and for the fifth harmonic it's 5. So to find the frequency of the third harmonic, I need to take the 105.4 hertz and multiply by 3. That gives me 316.2 hertz, and I can round that to two significant digits, 320 hertz. To find the next highest frequency, which is the fifth, I need to take the first harmonic and multiply by 5. I need to take 105.4 and multiply by 5, and that gives me 527.1 hertz. Again, I can round that to two significant digits, and it becomes 530 hertz.