Sources of Musical Sound — Question 2

Question 2

Adjacent resonance gaps identify different acoustic length scales

Tube A is open at both ends and has adjacent resonances at $300$ and $400\,\mathrm{Hz}$. Tube B is closed at one end and has adjacent resonances at $300$ and $500\,\mathrm{Hz}$. Use $c=340\,\mathrm{m/s}$ and neglect end corrections.

1. Infer both lengths and identify the observed harmonic numbers.

2. Explain why the larger spacing of tube B does not mean its fundamental is $200\,\mathrm{Hz}$.

Question 2 - Solution

Adjacent resonance gaps identify different acoustic length scales

(a) Method and calculation

For A, adjacent spacing is $c/(2L_A)=100\,\mathrm{Hz}$, so$$L_A=340/200=1.70\,\mathrm m,\qquad n=3,4.$$For B, adjacent allowed odd modes are separated by $c/(2L_B)=200\,\mathrm{Hz}$, so$$L_B=340/400=0.850\,\mathrm m,\qquad f_{B,1}=c/(4L_B)=100\,\mathrm{Hz}.$$

(b) Interpretation and further calculation

The B resonances are its third and fifth harmonics. Its sequence is$$f=(2m+1)f_{B,1},\quad\Delta f=2f_{B,1}.$$Thus adjacent-mode spacing equals the fundamental only for the complete integer sequence of an ideal open-open tube. The boundary condition must be known before interpreting a measured gap.

Physical check

Both tubes have the same fundamental but different allowed spectra and physical lengths.