18.Two wires are fixed on a sonometer.Their tensions are in the ratio 8:1, their lengths are in the ratio 36:35, the diameters are in the ratio 4:1 and densities are in the ratio 1:2.Find the frequencies of the beats produced if the note of the higher pitch has a frequency of 360/s.-------------

Expression for Fundamental frequency:
        $n=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$
Here, L is length the string, T is tension, and $\mu$ is mass per unit length.

Mass per unit length, $\mu =\frac{mass}{length}$
                                     $$\begin{gathered} =\frac{\rho \pi r^{2}L}{L} \\ =\pi r^2\rho \end{gathered}$$

Here, r is radius of the wire and $\rho$ is density of the material of the wire.

Hence, $n=\frac{1}{2L}\sqrt{\frac{T}{\pi r^2\rho }}$
             $=\frac{1}{2L}\sqrt{\frac{T}{\pi D^2\rho }}$

Here, D is diameter of the wire.

$$\begin{gathered} \frac{n_1}{n_2}=\left(\frac{L_2}{L_1}\right)\sqrt{\frac{T_1}{T_2}}\left(\frac{D_2}{D_1}\right)\sqrt{\frac{{\rho }_2}{{\rho }_1}} \\ =\left(\frac{35}{36}\right)\sqrt{\frac{8}{1}}\left(\frac{1}{4}\right)\left(\sqrt{\frac{2}{1}}\right) \\ =\frac{35}{36} \end{gathered}$$

Here, $n_2 > n_1$
The fequency of higher pitch is given as 360 Hz.

So,
 
$$\begin{gathered} \frac{n_1}{360}=\frac{35}{36} \\ n_1=350 Hz \end{gathered}$$

Frequency of beats:

$$\begin{gathered} n=n_2-n_1 \\ =360 - 350 \\ = 10 \mathrm{Hz} \end{gathered}$$