Question

Two wires are fixed in a sonometer. Their tensions are in the ratio $8:1$. The lengths are in the ratio $36:35$. The diameter are in the ratio $4:1$. Densities of the materials are in the ratio $1:2$. If the lower frequency in the setting is $360\text{Hz}$ The beat frequency, when the two wires are sounded together is

• A. 5 Hz
• B. 8 Hz
• C. 6 Hz
• D. 10 Hz

Answer 1

The correct option is D 10 Hz

Given, $\frac{T_1}{T_2}=\frac{8}{1},\frac{L_1}{L_2}=\frac{36}{35}$
$\frac{D_1}{D_2}=\frac{4}{1},\frac{{\rho }_1}{{\rho }_2}=\frac{1}{2}$

Let ${\mu }_1$ and ${\mu }_2$ be the linear mass densities, then
${\mu }_1=\pi \times \frac{D_1^2}{4}\times {\rho }_1$ and ${\mu }_2=\pi \times \frac{D_2^2}{4}\times {\rho }_2$

$\frac{{\mu }_1}{{\mu }_2}=(\frac{D_1}{D_2}{)}^2\times \frac{{\rho }_1}{{\rho }_2}=(\frac{4}{1}{)}^2\times \frac{1}{2}=\frac{8}{1}$
$:.\frac{f_1}{f_2}=\frac{L_2}{L_1}\times \sqrt{\frac{T_1}{T_2}\times \frac{{\mu }_2}{{\mu }_1}}=\frac{35}{36}\sqrt{\frac{8}{1}\times \frac{1}{8}}=\frac{35}{36}$

We know that $f_2>f_1$ and hence
we have $f_2=360$ and $f_1=350$ .

Having both of these terms, we can derive the value of beat frequency as
$f_2-f_1=10\text{Hz}$