Question

Two identical guitar strings are prepared such that they have the same length ( 0.65 m ) and are under the same amount of tension. The first string is plucked at one location, primarily exciting the second harmonic. The other string is plucked in a different location, primarily exciting the third harmonic. The resulting sounds give rise to a beat frequency of 378 Hz . What is the wave propagation speed on the guitar strings?

Answer 1

Answer 2

For a string fixed at both ends, the resonant frequencies are

$$f_n=\frac{nv}{2L}$$

where ?$v$ is the wave propagation speed, ?$L$ is the length of the string, and ?$n$ is the harmonic number. Since the propagation materials (the guitar strings) are identical, the harmonic number is the only difference between the two. When two sound waves with different frequencies combine, a phenomenon called "beats" can be heard (or at least observed on a plot of air pressure versus time in the case of large beat frequencies). The beat frequency is equal to the difference between the two frequencies being combined.

$$\begin{array}{rl}{f}_{\text{b}}& =\left|f_{\text{n}1}-f_{\text{n}2}\right|\\ \\ & =\left|\frac{n_{1}v}{2L}-\frac{n_{2}v}{2L}\right|\\ \\ & =\left|n_1-n_2\right|\frac{v}{2L}\end{array}$$

Solving this for the speed gives

?=2??b|?1−?2|=2(0.64 m)(351 Hz)(3−2)=450 m/s