Question

A guitar string is 90 cm long and has a mass of 3.7 g . The distance from the bridge to the support post is L=62cm, and the string is under a tension of 500 N . What are the frequencies of the fundamental and first two overtones?

Answer 1

To solve this problem we will apply the concept of frequency in a string from the nodes, the tension, the linear density and the length of the string, that is,

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

Here

n = Number of node

T = Tension

$\mu$ = Linear density

L = Length

Replacing the values in the frequency and value of n is one for fundamental overtone

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

$f = \frac{1}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})$

$\mathbf{f = 281.2Hz}$

Similarly plug in 2 for n for first overtone and determine the value of frequency

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

$f = \frac{2}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})$

$\mathbf{f = 562.4Hz}$

Similarly plug in 3 for n for first overtone and determine the value of frequency

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

$f = \frac{3}{2(62*10^{-2})}\bigg (\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}} \bigg)$

$\mathbf{f= 843.7Hz}$