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Two adjacent natural frequencies of an organ pipe are found to be550 Hz and 650 Hz. Calculate the fundamental frequency and length of this pipe .( Use v = 340 m//s.) |
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Answer» Solution :These will be fairly high harmonics of the pipe , which we except to be long pipe for a bass note . We use the WAVE under boundary conditions model . We are not TOLD in advance whether the pipe is open at one end or at both ends , so we MUST think about possibilities . Because harmonic frequencies are given by `f_(1) n` for open pipe , and `f_(1) ( 2n - 1)` for closed pipes , the difference between adjacent harmonics is constant in both cases . Therefore , we can find each harmonic below `650 Hz` by subtracting `Delta f_( Harmonic) = ( 650 Hz - 550 Hz ) = 100 Hz` from the previous value. The harmonic frequencies are the set of the following values : ltbgt `650 Hz , 550 Hz , 450 Hz , 350 Hz , 250 Hz , 150 Hz, 150 Hz and 50 Hz` a. So the fundamental frequency , is `50 Hz`. b. The wavelength of the fundamental vibration can be calculated from the speed of sound as ` LAMBDA = (v)/(f) = ( 340 m//s)/( 50.0 Hz) = 6.80 m` Because the step size `Delta f` is twice the fundamental frequency . We know the pipe is closed , with an ANTINODE at the open end , and a node at the closed end . The wavelength in this situation is four times the pipe length , so ` L = (lambda)/( 4) = 1.70 m` The frequency `50 Hz is a G` in low register . The harmonics measured in the problem are the ` 11 th and the 13 th`, which can be perfectly observable as resonances using a function generator , a small loudspeaker , and your ear as detector .For some ranks of organ pipes , higher harmonics like these can be important contributors to a brassy or ringing sound quality . |
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