Question: (II) A tight guitar string has a frequency of 540 Hz as its third harmonic. What will be its fundamental frequency if it is fingered at a length of only 70% of its original length?

In this problem, the fundamental frequency is evaluated by dividing the frequency of string with the number of harmonic.

The third harmonic frequency is \({f_3} = 540{\rm{ Hz}}\).

The fingered length is \({l_{{\rm{fingered}}}} = 0.7l\).

The fundamental frequency of the string is calculated as,

\(\begin{array}{c}{f_1} = \frac{{{f_3}}}{n}\\ = \frac{{540{\rm{ Hz}}}}{3}\\ = 180{\rm{ Hz}}\end{array}\)

The fundamental frequency of the fingered string is calculated as,

\(\begin{array}{l}{f_{{1_{{\rm{fingered}}}}}} = \frac{\nu }{{2{l_{{\rm{fingered}}}}}}\\{f_{{1_{{\rm{fingered}}}}}} = \frac{\nu }{{2 \times \left( {0.70l} \right)}}\end{array}\)

Here, v is the speed of sound.

Substitute the values in the above relation.

\(\begin{array}{c}{f_{{1_{{\rm{fingered}}}}}} = \frac{\nu }{{2 \times \left( {0.70l} \right)}}\\ = \frac{1}{{0.70}}{f_1}\\ = \frac{{180{\rm{ Hz}}}}{{0.70}}\\{f_{{1_{{\rm{fingered}}}}}} = 257.14{\rm{ Hz}}\end{array}\)

Thus, the fundamental frequency of fingered string is \(257.14{\rm{ Hz}}\).