Show that when a string fixed at its two ends vibrates in $1$ loop,$2$ loops,$3$ loops and $4$ loops,the frequencies are in the ratio $1 : 2 : 3 : 4$.

(N/A) For a string of length $L$ fixed at both ends,the frequency of the $n^{\text{th}}$ harmonic is given by the formula: $f_n = \frac{nv}{2L}$,where $n$ is the number of loops,$v$ is the speed of the transverse wave in the string,and $L$ is the length of the string.
Since $v$ and $L$ are constant for a given string under constant tension,the frequency $f_n$ is directly proportional to the number of loops $n$ $(f_n \propto n)$.
For $n=1, 2, 3, 4$,the frequencies are $f_1 = \frac{v}{2L}$,$f_2 = \frac{2v}{2L}$,$f_3 = \frac{3v}{2L}$,and $f_4 = \frac{4v}{2L}$.
Therefore,the ratio of the frequencies is $f_1 : f_2 : f_3 : f_4 = 1 : 2 : 3 : 4$.