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 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

• A. 5
• B. 8
• C. 6
• D. 10

Answer 1

The correct option is D

10

n = $\frac{1}{2l}$ $\sqrt{\frac{T}{n{r}^2\rho }}$= $\frac{1}{l}$ $\sqrt{\frac{T}{\pi d^2\rho }}$ (d = diameter of string)

$\Rightarrow$ $\frac{n_1}{n_2}$ = $\frac{l_2}{l_1}$ $\sqrt{\frac{T_1}{T_2}\times {\left(\frac{d_2}{d_1}\right)}^2\times \times \frac{{\rho }_2}{{\rho }_1}}$

= $\frac{35}{36}$ $\sqrt{\frac{8}{1}\times {\left(\frac{1}{4}\right)}^2\times \frac{2}{1}}$ = $\frac{35}{36}$ $\Rightarrow$ $n_1$ = $\frac{35}{36}$ $\times$ 360 = 350

Hence beat frequency = $n_1$ - $n_2$ = 10