Sources of Musical Sound — Question 5

Question 5

A string can be tuned to an air-column fundamental

A string of length $L_s=0.650\,\mathrm m$ and linear density $\mu=5.0\times10^{-4}\,\mathrm{kg/m}$ is fixed at both ends. It is tuned to the fundamental of a closed-open pipe of effective length $L_p=0.425\,\mathrm m$. Sound speed is $340\,\mathrm{m/s}$.

1. Find the common frequency and required string tension.

2. If the pipe effective length is halved, how must string tension change to match its new fundamental while string length and density remain fixed?

Question 5 - Solution

A string can be tuned to an air-column fundamental

(a) Method and calculation

The pipe sets the target frequency:$$f=\frac c{4L_p}=\frac{340}{1.70}=200\,\mathrm{Hz}.$$For the string $f=(2L_s)^{-1}\sqrt{F/\mu}$, giving$$F=\mu(2L_sf)^2=(5\times10^{-4})(260)^2=33.8\,\mathrm N.$$

(b) Interpretation and further calculation

Halving pipe length doubles its fundamental to $400\,\mathrm{Hz}$. For the string,$$\frac{F_2}{F_1}=\left(\frac{f_2}{f_1}\right)^2=4,\qquad F_2=135.2\,\mathrm N.$$The pipe and string change pitch through different physical controls: acoustic path length versus elastic-wave speed.

Physical check

The matched fundamental does not mean identical overtone spectra: an ideal string has all integer harmonics, while the closed-open pipe has odd ones.