A guitar string is 92 cm long and has a mass of 3.4 g. The distance from the bridge to the support post is \({\bf{l}} = {\bf{62}}{\rm{ }}{\bf{cm}}\), and the string is under a tension of 520 N. What are the frequencies of the fundamental and first two overtones?

The relation of the fundamental frequency with the tension and mass per unit length is applied in this problem.

The total length of the string is \({l_{{\rm{total}}}} = 92{\rm{ cm}}\).

The length from the bridge to the post is \({l_{{\rm{vibrating}}}} = 62{\rm{ cm}}\).

The tension in the string is \({F_{\rm{T}}} = 520{\rm{ N}}\).

The mass of the string is \(m = 3.4{\rm{ g}}\).

The fundamental frequency is calculated as;

\({f_1} = \frac{1}{{2{l_{{\rm{vibrating}}}}}}\sqrt {\frac{{{F_{\rm{T}}}}}{{m/{l_{{\rm{total}}}}}}} \)

Substitute the known values in the above equation.

\(\begin{aligned}{c}{f_1} &= \frac{1}{{2 \times \left( {62\;{\rm{cm}}\left( {\frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)} \right)}}\sqrt {\frac{{520{\rm{ N}}}}{{\frac{{\left( {3.4 \times {{10}^{ - 3}}{\rm{ kg}}} \right)}}{{92\;{\rm{cm}}\left( {\left( {\frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)} \right)}}}}} \\ &= 302.5{\rm{ Hz}}\end{aligned}\)

The second harmonic frequency is calculated as;

\(\begin{aligned}{c}{f_2} &= 2{f_1}\\ &= 2 \times 302.5{\rm{ Hz}}\\ &= 605{\rm{ Hz}}\end{aligned}\)

The third harmonic frequency is calculated as;

\(\begin{aligned}{c}{f_3} &= 3{f_1}\\ &= 3 \times 302.5{\rm{ Hz}}\\ &= 907.5{\rm{ Hz}}\end{aligned}\)

Thus, the fundamental, second harmonic and third harmonic frequencies are \(302.5{\rm{ Hz}}\), \(605{\rm{ Hz}}\) and \(907.5{\rm{ Hz}}\) respectively.