Answer:
(a) the fundamental frequency of this string is 65 Hz
(b) the harmonics of the given frequencies are third and fourth respectively.
(c) the length of the string is 2.74 m
Explanation:
Given;
mass density of the string, μ = 3 x 10⁻³ kg/m
tension of the string, T = 380 N
resonating frequencies, 195 Hz and 260 N
For the given resonant frequencies;
![195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3](https://tex.z-dn.net/?f=195%20%3D%20%5Cfrac%7Bn%7D%7B2l%7D%20%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cmu%7D%20%7D%20---%281%29%5C%5C%5C%5C260%20%3D%20%5Cfrac%7Bn%2B1%7D%7B2l%7D%20%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cmu%7D%20%7D%20---%282%29%5C%5C%5C%5Cdivide%20%5C%20%282%29%20%5C%20by%20%281%29%5C%5C%5C%5C%5Cfrac%7B260%7D%7B195%7D%20%3D%20%5Cfrac%7Bn%2B1%20%7D%7Bn%7D%20%5C%5C%5C%5C260n%20%3D%20195%28n%2B1%29%5C%5C%5C%5C260%20n%20%3D%20195%20n%20%2B%20195%5C%5C%5C%5C260n%20-%20195n%20%3D%20195%5C%5C%5C%5C65n%20%3D%20195%5C%5C%5C%5Cn%20%3D%20%5Cfrac%7B195%7D%7B65%7D%20%5C%5C%5C%5Cn%20%3D%203)
(c) From any of the equations, solve for Length of the string (L);
![195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m](https://tex.z-dn.net/?f=195%20%3D%20%5Cfrac%7Bn%7D%7B2l%7D%20%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cmu%7D%20%7D%20%5C%5C%5C%5C195%20%3D%20%5Cfrac%7B3%7D%7B2l%7D%5Csqrt%7B%5Cfrac%7B380%7D%7B3%5Ctimes%2010%5E%7B-3%7D%7D%20%7D%20%5C%5C%5C%5Cl%20%3D%20%5Cfrac%7B3%7D%7B2%5Ctimes%20195%7D%5Csqrt%7B%5Cfrac%7B380%7D%7B3%5Ctimes%2010%5E%7B-3%7D%7D%20%7D%5C%5C%5C%5Cl%20%3D%202.74%20%5C%20m)
(a) the fundamental frequency is calculated as;
![f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o = 65 \ Hz](https://tex.z-dn.net/?f=f_o%20%3D%20%5Cfrac%7B1%7D%7B2l%7D%20%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cmu%7D%20%7D%20%5C%5C%5C%5Cf_o%20%3D%20%5Cfrac%7B1%7D%7B2%5Ctimes%202.74%7D%20%5Csqrt%7B%5Cfrac%7B380%7D%7B3%5Ctimes%2010%5E%7B-3%7D%20%7D%20%7D%5C%5C%5C%5Cf_o%20%3D%20%2065%20%5C%20Hz)
(b) harmonics of the given frequencies;
the first harmonic (n = 1) = f₀ = 65 Hz
the second harmonic (n = 2) = 2f₀ = 130 Hz
the third harmonic (n = 3) = 3f₀ = 195 Hz
the fourth harmonic (n = 4) = 4f₀ = 260 Hz
Thus, the harmonics of the given frequencies are third and fourth respectively.