Question

Two identical guitar strings are stretched with the same tension between supports that are not the same distance apart. The fundamental frequency of the higher-pitched string is 380Hz, and the speed of transverse waves in both wires is 200 m/s. How much longer is the lower-pitched string if the beat frequency is 6Hz?

Answer 1

The velocity of the wave in the string is

The velocity of the wave in the string is v = f lambda lambda = v/f lambda = 200/380 The wavelength of the wave for the first string is lambda = 200/380 = 0.526 m The length of the string is l = lambda/2 = 0.263 m the frequency of the second string is f = 380 - 6 = 374Hz The wave length is lambda prime = 200/374 = 0.535 m the length of the string is l prime = lambda prime/2 = 0.535/2 = 0.267 m 0.267 - 0.263 = 0.004m is the change in length in the both the strings.

$\lambda =\frac{v}{f}$

   $\lambda =\frac{200}{380}$

The wavelength of the wave for the first string is $\lambda =\frac{200}{380}= 0.526m$

The lenght of the strng is $l= \frac{\lambda }{2}=0.263m$

the frequency of the second string is f' = 380-6 = 374Hz

The wave length is $\lambda ' = \frac{200}{374} = 0.535m$

the length of the string is l' =    $\frac{\lambda ' }{2}= \frac{0.535}{2}=0.267m$   

0.267-0.263 = 0.004m is the change in length in the both the strings.