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Unformatted text preview: Physics 1240 Homework 4 solutions Brief solutions to homework set #4. Your numbers will be different, but the idea is the same. I’m just using sample numbers here. 1. Musical notes have names (A,B,C, etc). The lowest A on a piano is called A0. If you go up an octave, the next A would be called A1. An octave higher, A2. The A near the middle of the keyboard is A4 (also called “concert A”). A similar pattern is true for all the notes. See the picture in the back flyleaf of your textbook for details! In a certain set of organ pipes, the speaking length (that’s an effective physical length) for the pipe that makes the note C5 (f=524 Hz) is 32.82 cm. How long a pipe do you expect for C2? What is the wavelength of sound produced by the C5 pipe described in this problem? When you write this problem up, start by briefly explaining this calculation. It’s not the same as the length of the pipe, it should be different by a factor of 2 (that’s not an accident, we’ll discuss why later). The longest organ pipe in the world (according to my quick web search) is 64 feet long. Based on the numbers in this problem, could you actually hear the sound produced by this pipe? Why or why not? What might be the effect of such a pipe? What would you need to do (lengthen? shorten? by how much?) to this pipe to get a sound with a wavelength twice as long? The note C2 is 3 octaves lower than C5, so its frequency is 65.5 Hz. This requires making the length of the pipe 8 times longer (doubling the length of the pipe 3 times. Therefore the pipe length must be 8 × . 3282 m ≈ 2 . 625 m. The wavelength associated with this note is λ = v/f = 344 m/s/65.5 Hz ≈ 5 . 25 m ≈ 15 . 75 feet....
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- Spring '08
- Work, Wavelength, 5.25 m, 0.3282 m, 2.625 m