Sources of Musical Sound — Question 4

Question 4

Thermal tuning and length tuning act through different factors

An ideal pipe has fundamental $440\,\mathrm{Hz}$ at $293\,\mathrm K$. At fixed composition, $c\propto\sqrt T$. Neglect the thermal expansion of the pipe. It warms uniformly to $303\,\mathrm K$.

1. Find the new fundamental frequency.

2. By what factor must its effective length change to restore $440\,\mathrm{Hz}$ at the new temperature? For an original effective length $0.390\,\mathrm m$, find the extension.

Question 4 - Solution

Thermal tuning and length tuning act through different factors

(a) Method and calculation

For a fixed boundary type, $f\propto c/L$. With unchanged length,$$\frac{f_2}{f_1}=\sqrt{\frac{303}{293}}=1.016922,\qquad f_2=447.446\,\mathrm{Hz}.$$The pipe becomes sharper because the sound speed increases.

(b) Interpretation and further calculation

To keep frequency constant, $L_2/L_1=c_2/c_1$, hence$$\Delta L=L_1\left(\sqrt{303/293}-1\right)=0.00660\,\mathrm m.$$The pipe must lengthen by about $6.60\,\mathrm{mm}$. Shortening would raise the pitch further instead of compensating for warming.

Physical check

The length correction is about $1.69\%$, approximately half the fractional absolute-temperature change for this small temperature interval.