In Fig. 16-43, an aluminum wire, of length ${\mathit{L}}_{\mathbf{l}}\mathbf{=}\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$, cross$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathit{c}{\mathit{m}}^{\mathbf{2}}$-sectional area, and density$\mathbf{2}\mathbf{.}\mathbf{60}\mathbf{}\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$, is joined to a steel wire, of densityrole="math" localid="1661166375385" $\mathbf{7}\mathbf{.}\mathbf{80}\mathbf{}\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$and the same cross-sectional area. The compound wire, loaded with a block of mass$\mathit{m}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathit{k}\mathit{g}$, is arranged so that the distance${\mathit{L}}_{\mathbf{2}}$from the joint to the supporting pulley is 86.6 cm. Transverse waves are set up on the wire by an external source of variable frequency; a node is located at the pulley. (a) Find the lowest frequency that generates a standing wave having the joint as one of the nodes. (b) How many nodes are observed at this frequency?

The length of aluminum wire,$L_1=60.0cmor0.6m$

The length of steel wire,$L_2=86.6cmor0.866m$

Cross-sectional area of both wires,$A_1=A_2=1.00\times {10}^{-2}c{m}^{2}or1.00\times {10}^{-6}{m}^2$

Density of aluminum wire,${\rho }_1=2.60\frac{g}{c{m}^3}or2.6\times {10}^3\frac{kg}{m^3}$

Density of steel wire,${\rho }_2=7.80\frac{g}{c{m}^3}or7.8\times {10}^3\frac{kg}{m^3}$

Mass of the block,$m=10.0kg$

From the free body diagram of the block, we can find the tension in the string. We are given enough information to calculate the mass densities of each wire. Using mass density and tension, we can find the velocity of the wave generated on these strings. Now as the same wave is generated on both the strings, using the above parameters, we can find the wavelength. Once we get the wavelength, we can find the frequency and nodes generated.

Formula:

The linear density of the string,$\mu =\frac{m}{L}..........\left(1\right)$

The mass of a body in terms of density,$m=\rho \times Volume.............\left(2\right)$

The volume of a body,$Area\times Length..............\left(3\right)$

The velocity of a body,$v=\sqrt{\frac{T}{\mu }}.............\left(4\right)$

The velocity of a wave, $v=\lambda \times f............\left(5\right)$

Free body diagram of the block,

So we can say that the tension of a body is:

$$\begin{gathered} T=mg \\ =10\times 9.8 \\ =98N \end{gathered}$$

So, from equation (1), we get that

${\mu }_1=\frac{m_1}{L_1}and{\mu }_2=\frac{m_2}{L_2}$

Using equations (2) and (3), we get the masses of the wires as given:

$$\begin{gathered} m_1={\rho }_1\times A_1\times L_1 \\ {m}_2={\rho }_2\times A_2\times L_2 \end{gathered}$$

Substituting the values of masses in equation (1), we get the linear density of both the wires as given:

$$\begin{gathered} {\mu }_1=\frac{{\rho }_1\times A_1\times L_1}{L_1} \\ ={\rho }_1\times A_1 \\ =\left(2.6\times {10}^3\right)\times \left(1\times {10}^{-6}\right) \\ =2.6\times {10}^{-3}kg/m \\ {\mu }_2=\frac{{\rho }_2\times A_2\times L_2}{L_2} \\ ={\rho }_2\times A_2 \\ =\left(7.8\times {10}^3\right)\times \left(1\times {10}^{-6}\right) \\ =7.8\times {10}^{-3}kg/m \end{gathered}$$

Substituting values of tension and mass of aluminum wire in equation (4), we get the velocity as:

$$\begin{gathered} v_1=\sqrt{\frac{98}{2.6\times {10}^{-3}}} \\ =194.195m/s \end{gathered}$$

Substituting values of tension and mass of steel wire in equation (4), we get the velocity as:

$$\begin{gathered} v_2=\sqrt{\frac{98}{7.8\times {10}^{-3}}} \\ =112.089m/s \end{gathered}$$

Again, by using equation (5), we get the ratios of velocity as given:

$$\begin{gathered} \frac{v_2}{v_1}=\frac{{\lambda }_2\times f}{{\lambda }_1\times f}(\sin ce.thefrequencyofoscillationissame) \\ \frac{{\lambda }_2}{{\lambda }_1}=\frac{112.089}{194.195} \\ =0.577 \end{gathered}$$

Length of the wires representing the position of n^th node of the standing wave can be given as:

localid="1661167705861" $$\begin{gathered} L_1=n_1\left(\frac{{\lambda }_1}{2}\right)...........\left(6\right) \\ {L}_2=n_2\left(\frac{{\lambda }_2}{2}\right)...........\left(7\right) \end{gathered}$$

Hence, the ration of the lengths of the wires can be given using equations (6) and (7) as:

$$\begin{gathered} \frac{L_2}{L_1}=\frac{n_2\left({\displaystyle \frac{{\lambda }_2}{2}}\right)}{n_1\left(\frac{{\lambda }_1}{2}\right)} \\ \frac{L_2}{L_1}=\left(\frac{n_2}{n_1}\right)\times \left(\frac{{\lambda }_2}{{\lambda }_1}\right) \\ \frac{0.866}{0.6}=\left(\frac{n_2}{n_1}\right)\times 0.577 \\ \left(\frac{n_2}{n_1}\right)=\frac{5}{2} \end{gathered}$$

From this we can say that,

The lowest values of nodes $n_1=2$and$n_2=5$.

Hence, from equation (6), the wavelength of aluminum wire can be given as:

$$\begin{gathered} {\lambda }_1=\frac{2L_1}{n_1} \\ =\frac{2\times 0.6}{2} \\ =0.6m \end{gathered}$$

Hence, the value of frequency of aluminum wire is given as:

$$\begin{gathered} f=\frac{v_1}{{\lambda }_1} \\ =\frac{194.195}{0.6} \\ =323.6Hz \\ \approx 324Hz \end{gathered}$$

Hence, the value of the lowest frequency where standing node is observed is 324 Hz

At the frequency of 324 Hz , the number of antinodes is

$$\begin{gathered} Antinodes=n_1+n_2 \\ =2+5 \\ =7 \end{gathered}$$

$$\begin{gathered} Totalnumberofnodes=\left(numberofantinodes\right)+1 \\ =7+1 \\ =8 \end{gathered}$$

Hence, the total number of standing nodes is 8.