Answers to Chapter 12 Homework

1. (a) We find the speed of sound at 15°C:

                        v = (331 + 0.60T) m/s = [331 + (0.60)(15°C)] m/s = 340m/s

          The wavelength of the fundamental frequency is l₁ = 4L. we find the length from

                       v = l₁f₁ = 4Lf₁;

                      340m/s = 4L(294Hz), which gives L = 0.289m.

     (b) For helium we have

                      v =l₁f₁ = 4Lf₁;

                     1005m/s = 4(0.289m)f₁, which gives f₁ = 869Hz.

          Note that we have no correction for the 5°C temperature change.

2. (a) For an open pipe all harmonics are present, the difference in frequencies is the fundamental frequency, and all frequencies will be integral multiples of the difference. For a closed pipe only odd harmonics are present, the difference in frequencies is twice the fundamental frequency, and frequencies will not be integral multiples of the difference but odd multiples of half the difference. for this pipe we have

                      Df =280Hz - 240Hz = 40Hz.

      Because we have frequencies that are integral multiples of this, the pipe is   Open,  witha fundamental frequency of 40Hz.

      The wavelength of the fundamental frequency is l₁ = 2L. We find the length from

                      v = l₁f₁ = 2Lf₁;

                     343m/s = 2L(40Hz), which gives L = 4.3m.