The shortest pipes used in organs are about 7.9 cm long. (a) Estimate the fundam

Question

The shortest pipes used in organs are about 7.9 cm long. (a) Estimate the fundamental frequency of a pipe this longthat is open at both ends.
(b) For such a pipe, estimate the harmonic number n of thehighest-frequency harmonic that is within the audible range. (Theaudible range of human hearing is about 20 to 20,000 Hz.)
The shortest pipes used in organs are about 7.9 cm long. (a) Estimate the fundamental frequency of a pipe this longthat is open at both ends.
(b) For such a pipe, estimate the harmonic number n of thehighest-frequency harmonic that is within the audible range. (Theaudible range of human hearing is about 20 to 20,000 Hz.)

Explanation / Answer

length L = 7.9 cm =0.079 m (a). Fundamental frequency f = v / 2L                                             = 343 / ( 2 * 0.079 )                                             = 2170.88 Hz (b). For highest frequency F = 20000 Hz                                     nf = 20000 Hz           Highestharmonic number   n = 20000 Hz / f                                                       =9.2128 SO, Required n = 9

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