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,
[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]
Here
n = Number of node
T = Tension
[tex]\mu[/tex] = Linear density
L = Length
Replacing the values in the frequency and value of n is one for fundamental overtone
[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]
[tex]f = \frac{1}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})[/tex]
[tex]\mathbf{f = 281.2Hz}[/tex]
Similarly plug in 2 for n for first overtone and determine the value of frequency
[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]
[tex]f = \frac{2}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})[/tex]
[tex]\mathbf{f = 562.4Hz}[/tex]
Similarly plug in 3 for n for first overtone and determine the value of frequency
[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]
[tex]f = \frac{3}{2(62*10^{-2})}\bigg (\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}} \bigg)[/tex]
[tex]\mathbf{f= 843.7Hz}[/tex]
