Question

Two identical guitar strings are prepared such that they have the same length ( 0.66 m ) and are under the same amount of tension. The first string is plucked at one location, primarily exciting the fourth harmonic. The other string is plucked in a different location, primarily exciting the second harmonic. The resulting sounds give rise to a beat frequency of 351 Hz . What is the wave propagation speed on the guitar strings

Answer 1

`154.44\ \text{m/s}`

Step-by-step explanation:

L = Length of guitar string = 0.66 m

Difference of fourth and first harmonic = 351 Hz

v = Wave propagation speed on the guitar strings

So,

`(4v)/(2L)-(v)/(2L)=351\\\Rightarrow (3v)/(2L)=351\\\Rightarrow v=(351* 2* 0.66)/(3)\\\Rightarrow v=154.44\ \text{m/s}`

The wave propagation speed on the guitar strings is
`154.44\ \text{m/s}`.