91Ó°ÊÓ

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,$A_1=A_2=1.00×{10}^{-2}c{m}^{2}or1.00×{10}^{-6}{m}^2$

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

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

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,$\mathrm{μ}=\frac{m}{L}..........\left(1\right)$

The mass of a body in terms of density,$m=\mathrm{ÒÏ}×Volume.............\left(2\right)$

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

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

The velocity of a wave, $v=\mathrm{λ}×f............\left(5\right)$

Free body diagram of the block,

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

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

So, from equation (1), we get that

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

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

$$\begin{gathered} m_1={\mathrm{ÒÏ}}_1×A_1×L_1 \\ {m}_2={\mathrm{ÒÏ}}_2×A_2×L_2 \end{gathered}$$

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

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

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

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

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

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

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

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

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

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

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

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

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:

$$\begin{gathered} {\mathrm{λ}}_1=\frac{2L_1}{n_1} \\ =\frac{2×0.6}{2} \\ =0.6m \end{gathered}$$

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

$$\begin{gathered} f=\frac{v_1}{{\mathrm{λ}}_1} \\ =\frac{194.195}{0.6} \\ =323.6Hz \\ ≈324Hz \end{gathered}$$

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.