Question

Two identical guitar strings are prepared such that they have the same length (0.67 m) and are under the same amount of tension. The first string is plucked at one location, primarily exciting the fifth harmonic. The other string is plucked in a different location, primarily exciting the fourth harmonic. The resulting sounds give rise to a beat frequency of 3.60 x 102 Hz What is the wave propagation speed on the guitar strings? m/s 241.2 wave propagation speed:

Answer 1

We know that the wavelength of nth harmonics $\lambda$_n = 2L/n, where, L is the length of the string

Now, length of the string is 0.67m

$\therefore$ The wavelength of 5th harmonics $\lambda$5  = 2*0.67/5 = 0.268m

The wavelength of 4th harmonics $\lambda$4  = 2*0.67/4 = 0.335m

Now, beat frequency f₅ - f₄ = 3.6 * 10² Hz = 360 Hz (1) (f₅ and f₄ are the frequencies of two respectively)

Let, the velocity of the propagated wave is V m/s

Then, from (1), V(1/$\lambda$5 - 1/$\lambda$4) = 360

=> V(3.73 - 2.99) = 360

=> V = 360/0.74

=> V = 486.5 m/s

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