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Two uniform strings $A$ and $B$ made of steel are made to vibrate under the same tension. if the first overtone of $ A$ is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B,$ the ratio of the lengths of the strings is
$1: 2$
$1:3$
$1:4$
$1:6$
Solution
(b) First overtone of string $A$= Second overtone of string $B.$
$ \Rightarrow $ Second harmonic of $A$ = Third harmonic of $B$
$ \Rightarrow $ ${n_2} = {n_3}$$ \Rightarrow $ ${\left[ {2({n_1})} \right]_A} = {\left[ {3({n_1})} \right]_B}$ ( $\because$ ${n_1} = \frac{1}{{2l}}\sqrt {\frac{T}{{\pi {r^2}\rho }}} $)
==>$2\,\left[ {\frac{1}{{2{l_A}{r_A}}}\sqrt {\frac{T}{{\pi \rho }}} } \right] = 3\,\left[ {\frac{1}{{2{l_B}{r_B}}}\sqrt {\frac{T}{{\pi \rho }}} } \right]$
$\frac{{{l_A}}}{{{l_B}}} = \frac{2}{3}\frac{{{r_B}}}{{{r_A}}} \Rightarrow \frac{{{l_A}}}{{{l_B}}} = \frac{2}{3} \times \frac{{{r_B}}}{{(2{r_B})}} = \frac{1}{3}$
