Sources of Musical Sound — Question 8

Question 8

An ideal open tone hole shortens the active air column

A tube is closed at $x=0$ and open at physical length $0.80\,\mathrm m$. An ideal large side hole can impose a pressure node at $x=0.50\,\mathrm m$. For this model, ignore the distal segment after opening the hole, end corrections and losses; use $c=340\,\mathrm{m/s}$.

1. Find the fundamental with the hole closed and with it open.

2. Find the third-harmonic frequency in each state and explain why a small real hole need not impose this ideal result.

Question 8 - Solution

An ideal open tone hole shortens the active air column

(a) Method and calculation

With the hole closed, the active closed-open length is $0.80\,\mathrm m$; with it open, it is $0.50\,\mathrm m$. Thus$$f_{\mathrm{closed\ hole}}=340/[4(0.80)]=106.25\,\mathrm{Hz},$$$$f_{\mathrm{open\ hole}}=340/[4(0.50)]=170.0\,\mathrm{Hz}.$$

(b) Interpretation and further calculation

The odd-harmonic sequence gives third harmonics $318.75$ and $510.0\,\mathrm{Hz}$ respectively. All corresponding frequencies rise by $0.80/0.50=1.60$ in the ideal model. A small hole has finite acoustic impedance and need not force zero pressure; the distal air column and radiation can then affect tuning.

Physical check

The hole changes a boundary condition, not the speed of sound in the unchanged gas.