A wire of length L, mass per unit length mu, and tension T is plucked. The ratio of the frequency of the second harmonic to the fundamental frequency…

For a stretched string fixed at both ends, the frequencies of harmonics are given by $f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$, where $n = 1, 2, 3, \dots$. The fundamental frequency ($n=1$) is $f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, and the frequency of the second harmonic ($n=2$) is $f_2 = \frac{2}{2L} \sqrt{\frac{T}{\mu}} = 2f_1$. Thus, the ratio $f_2 : f_1$ is $2 : 1$.