Preview only show first 10 pages with watermark. For full document please download

Halliday 2 Cap 18 - P18 065

Inserir Descrição


  • Rating

  • Date

    December 2018
  • Size

  • Views

  • Categories



65. The points and the least-squares fit is shown in the graph below. The graph has frequency in Hertz along the vertical axis and 1/L in inverse meters along the horizontal axis. The function found by the least squares fit procedure is f = 276(1/L) + 0.037. Assuming this fits either the model of an open organ pipe (mathematically similar to a string fixed at both ends) or that of a pipe closed at one end, as discussed in the textbook, then f = v/2L in the former case or f = v/4L in the latter. Thus, if the least-squares slope of 276 fits the first model, then a value of v = 2(276) = 553 m/s is implied. In the second model (the pipe with only one end open) we find v = 4(276) = 1106 m/s which is more “in the ballpark” of the 1400 m/s value cited in the problem. This suggests that the acoustic resonance involved in this situation is more closely related to the n = 1 case of Figure 18-15(b) than to Figure 18-14. 40 35 30 25 frequency 20 15 10 5 0.02 0.04 0.06 0.08 reciprocal_L 0.1 0.12