Sources of Musical Sound — Question 10

Question 10

Two spectral lines can identify a boundary type without fixing length

An ideal uniform tube is either open-open or closed-open. It has resonances at exactly $200$ and $300\,\mathrm{Hz}$; they are not stated to be adjacent. Sound speed is $340\,\mathrm{m/s}$. Ignore end corrections.

1. Show which boundary type is impossible.

2. Find the possible open-open fundamentals and lengths. What additional observation selects the shortest allowed length?

Question 10 - Solution

Two spectral lines can identify a boundary type without fixing length

(a) Method and calculation

A closed-open tube has only odd multiples of its fundamental. The observed ratio $300/200=3/2$ would require$$\frac{2m+1}{2n+1}=\frac32\Rightarrow2(2m+1)=3(2n+1).$$The left side is even and the right side odd, a contradiction. Therefore the tube must be open-open under the stated alternatives.

(b) Interpretation and further calculation

For open-open modes, $200=nf_0$ and $300=mf_0$, so $n=2j$, $m=3j$ with positive integer $j$. Therefore$$f_0=100/j\,\mathrm{Hz},\qquad L=\frac{340}{2f_0}=1.70j\,\mathrm m.$$If the two observed modes are adjacent, $m-n=j=1$, selecting $L=1.70\,\mathrm m$. Without that information longer lengths remain possible.

Physical check

Frequency ratios can exclude a spectrum while leaving its overall mode numbering ambiguous.