Q28E - University Physics with Modern Physics

Question

The Vocal Tract. Many opera singers (and some pop singers) have a range of about ${\mathbf{2}}^{\mathbf{1}\mathbf{/}\mathbf{2}}$ octaves or even greater. Suppose a soprano’s range extends from A below middle C (frequency 220 Hz) up to E-flat above high C (frequency 1244 Hz). Although the vocal tract is quite complicated, we can model it as a resonating air column, like an organ pipe, that is open at the top and closed at the bottom. The column extends from the mouth down to the diaphragm in the chest cavity, and we can also assume that the lowest note is the fundamental. How long is this column of air if v = 354 m/s? Does your result seem reasonable, on the basis of observations of your own body?

Step-by-Step Solution

Answer

The length of the air column in the vocal tract is 0.39 m.

STEP 1 Concept of the frequency of standing wave in a stopped pipe

A pipe which has a closed end is given by 

$\mathbf{f}\mathbf{=}\mathbf{n}\frac{\mathbf{v}}{\mathbf{4}\mathbf{L}}$ 

where, f is the frequency of nth harmonic, v is the velocity of the wave, n represents the number of harmonic (n — 1, 3, 5, ...), L is the length of the pipe. 

Step 2 Calculate the length of the column

The frequency of the $n^{th}$ harmonic is given by the formula-

$$\begin{gathered} \mathrm{f}=\mathrm{n}\frac{\mathrm{v}}{4\mathrm{L}} \\ \mathrm{L}=\frac{\mathrm{nv}}{4\mathrm{f}} \\ =\frac{1\times 344\mathrm{m}/\mathrm{s}}{4\times 220 \mathrm{Hz}} \\ =0.39 \mathrm{m} \end{gathered}$$

Therefore, the length of the column is 0.39 m.