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

Answer:

$154.44\ \text{m/s}$

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,

$\dfrac{4v}{2L}-\dfrac{v}{2L}=351\\\Rightarrow \dfrac{3v}{2L}=351\\\Rightarrow v=\dfrac{351\times 2\times 0.66}{3}\\\Rightarrow v=154.44\ \text{m/s}$

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