A clarinet can be treated as a resonant tube with one closed end. The fundamental frequency is 147 Hz. I was able to figure out the length of the clarinet: it is .583 m BUt the question is to calculate the frequencies of the first three overtones. Would the overtone equatinos used to find the wavelength be (wave) = 4(length) (wave) = 2 (length) wave = length Or am I completly off track?
Here's a hint: For the fundamental frequency, the length of the tube equals 1/4 of a wavelength. For the first overtone (the next allowable frequency), how many wavelengths fit into the tube?
How does that work? If the fundamental frequency is 147 Hz then (147/4) equals the length? =36.75 this is not what I found..... Do I divide by four then?
147 Hz is the frequency, not the wavelength! Use the wave equation ([itex]v = f \lambda[/itex]) to find the wavelength, if you want to. But you don't need to find the wavelength or the length of the tube to solve this problem. What I want you to understand is the relationship between the wavelength of the fundamental frequency and the wavelength of the first harmonic. (If you know how the wavelengths relate, then you should be able to figure out how the frequencies relate via that wave equation.) Draw yourself a picture of the open/closed tube and draw the fundamental wave.