Answer:
[tex]3.93\times10^-^3kg.m^-^1[/tex]
Explanation:
The length,[tex]L[/tex] of the string , fundamental frequency [tex](f)[/tex] and the tension [tex](F)[/tex] on the string are related as:
[tex]L=\frac{1}{2f_1}\sqrt{\frac{F}{(m/L)}}\\\\\frac{2L}{\sqrt F}=\frac{1}{f_1\sqrt{(m/L)}}\\[/tex]
#Since both E and G have the same length and tension on them:
[tex]\frac{1}{f_1_,_G\sqrt{(m/L)_G}}=\frac{1}{f_1_,_E\sqrt{(m/L)_E}}[/tex]
Where [tex](m/L)_i[/tex] are the linear densities, [tex]f_1_,_i[/tex] the fundamental frequencies.
#taking square and inverse on both sides, we have:
[tex]f^2_1_,_G(m/L)_G=f^2_1_,_E(m/L)_E\\\\(m/L)_G=\frac{f^2_1_,_E}{f^2_1_,_G}(m/L)_E\\\\(m/L)_G=\frac{(659.3^2)}{(196^2)}\times 3.47\times 10^-^4=3.93\times10^-^3kg.m^-^1[/tex]
Hence, the linear density of the G string is 0.00393kg/m
