Sources of Musical Sound — Question 3

Question 3

End corrections prevent an exact octave after closing one end

A tube has physical length $L=0.500\,\mathrm m$ and radius $r=0.010\,\mathrm m$. Use an end correction $e=0.60r$ for each open end and zero correction at a rigid closure. Take $c=340\,\mathrm{m/s}$.

1. Find the fundamental when both ends are open and when one end is closed.

2. Find their frequency ratio and compare with the ideal ratio of two. Explain the direction of the discrepancy.

Question 3 - Solution

End corrections prevent an exact octave after closing one end

(a) Method and calculation

The effective lengths are $L_{oo}=L+2e=0.512\,\mathrm m$ and $L_{co}=L+e=0.506\,\mathrm m$. Therefore$$f_{oo}=\frac{340}{2(0.512)}=332.031\,\mathrm{Hz},\qquad f_{co}=\frac{340}{4(0.506)}=167.984\,\mathrm{Hz}.$$

(b) Interpretation and further calculation

The ratio is$$\frac{f_{oo}}{f_{co}}=\frac{2(L+e)}{L+2e}=1.97656<2.$$Closing one end removes one end correction as well as changing the boundary condition. The closed-open effective length is slightly shorter, making its fundamental slightly higher than exactly half the open-open frequency.

Physical check

As $e/L\to0$, the ratio approaches two. The correction coefficient is supplied rather than treated as universal for every opening geometry.