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Two wires are fixed in a sonometer. Their tensions are in the ratio $8 : 1$. The lengths are in the ratio $36:35.$ The diameters are in the ratio $4 : 1$. Densities of the materials are in the ratio $1 : 2$. If the lower frequency in the setting is $360 Hz.$ the beat frequency when the two wires are sounded together is
$5$
$8$
$6$
$10$
Solution
(d) Frequency in a stretched string is given by
$n = \frac{1}{{2l}}\sqrt {\frac{T}{{\pi {r^2}\rho }}} = \frac{1}{l}\sqrt {\frac{T}{{\pi {d^2}\rho }}} $ ($d =$ Diameter of string)
==> $\frac{{{n_1}}}{{{n_2}}} = \frac{{{l_2}}}{{{l_1}}}\sqrt {\frac{{{T_1}}}{{{T_2}}} \times {{\left( {\frac{{{d_2}}}{{{d_1}}}} \right)}^2} \times \left( {\frac{{{\rho _2}}}{{{\rho _1}}}} \right)} $
$ = \frac{{35}}{{36}}\sqrt {\frac{8}{1} \times {{\left( {\frac{1}{4}} \right)}^2} \times \frac{2}{1}} = \frac{{35}}{{36}}$$ \Rightarrow {n_2} = \frac{{36}}{{35}} \times 360 = 370$
Hence beat frequency = ${n_2} – {n_1} = 10$
