A sonometer wire has a length of $114 \mathrm{~cm}$, between two fixed ends. Where should two bridges be…
A sonometer wire has a length of $114 \mathrm{~cm}$, between two fixed ends. Where should two bridges be placed so as to divide the wire into three segments (in $\mathrm{cm}$ ) whose fundamental frequencies are in the ratio $1: 3: 4$ ?
$l_1, l_2, l_3=18,24,72$
$l_1, l_2, l_3=24,18,72$
$l_1, l_2, l_3=72,18,24$
$l_1, l_2, l_3=72,24,18$
Solution
If length of the wire between the two bridges is $l$, the frequency of vibration
$n=\frac{1}{2 l} \sqrt{\frac{T}{m}}$
$\therefore \quad n \propto \frac{1}{l}$
Therefore, $n_1: n_2: n_3=\frac{1}{l_1}: \frac{1}{l_2}: \frac{1}{l_3}$
$n_1: n_2: n_3=\frac{1}{\frac{114 \times 1}{8}}: \frac{1}{\frac{114 \times 3}{8}}: \frac{1}{\frac{114 \times 4}{8}}$
or $\quad n_1: n_2: n_3=8: \frac{8}{3}: 2$
or $\quad n_1: n_2: n_3=72: 24: 18$