A musician blows into an open pipe & produces its fundamental frequency. The pipe is 0.70 meters long. (Use velocity of sound = 343 m/s)
(a) What is the fundamental frequency?
(b) If he wanted to play a higher note, what would be the next higher frequency that can be played on this pipe?
(c) How many wavelengths fit inside the tube in this case?
(d) If the musician were to replace his open pipe with a closed pipe, what is the one possible length for the pipe if he wants to produce the same (higher) frequency?
any help would be greatly appreciated!!! thank u soooooo much!!! :)
2007-02-19 08:39:24 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
a) F1 = u/(2L) = 343/(2*.7) = 245 Hz
b) F2 = 2F1 = 490 Hz
c) λ = L, so 1.0 wavelength fits in the pipe
d) L = u/4F2 = .175 m
2007-02-19 08:59:16 · answer #1 · answered by Steve 7 · 1⤊ 0⤋
(a)343/(2x0.70) Hz (b) twice the fundamental (c)half in fundamental and one in next called first overtone (d)1/4
2007-02-27 10:32:50 · answer #2 · answered by ukmudgal 6 · 0⤊ 0⤋