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Two identical guitar strings are prepared such that they have the same length (2.17 m) and are under the same amount of tension. The first string is plucked at one location, primarily exciting the 5th harmonic. The other string is plucked in a different location, primarily exciting the 4th harmonic. The resulting sounds give rise to a beat frequency of 351 Hz. What is the wave propagation speed on the guitar strings?

Respuesta :

Answer:

[tex]v = 1523.3 m/s[/tex]

Explanation:

For two guitars we have the situation given here

First guitar is in 5th harmonic

so we have

[tex]5 \frac{\lambda}{2} = 2.17 m[/tex]

[tex]\lambda_1 = 0.868 m[/tex]

now we have

[tex]f_1 = \frac{v}{\lambda_1}[/tex]

now for second guitar we have

[tex]4 \frac{\lambda}{2} = 2.17[/tex]

[tex]\lambda_2 = 1.085 m[/tex]

now for second guitar we have

[tex]f_2 = \frac{v}{\lambda_2}[/tex]

now we know that

[tex]f_1 - f_2 = f_{beat}[/tex]

[tex]\frac{v}{0.868} - \frac{v}{1.085} = 351[/tex]

[tex]v(0.23) = 351 [/tex]

[tex]v = 1523.3 m/s[/tex]

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