Question by Shawn: How many overtones are present within the audible range for a 2.53 m long organ pipe at 10°C if it is open?
How many overtones are present within the audible range for a 2.53 m long organ pipe at 10°C if it is open?
How many are present if it is closed?
Best answer:
Answer by kirchwey
A formula for sound speed is c = 331.3+0.606*T(C) (see ref.), so we'll set c = 331.3+6.06 = 337.36 m/s. Then the fundamental frequency of the pipe = 337.36/(2*2.53) = 66.6719 Hz. Each integer multiple of this frequency can be produced by the pipe. Thus the number of audible overtones including the fundamental = INT(20000/66.6719) = 299.
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