To solve the problem step by step, we need to find the lowest frequency of excitation for which standing waves are formed such that the joint in the wire is a node. We will also determine the total number of nodes at this frequency.
### Step 1: Determine the Lengths of the Wires
The total length of the compound wire is given as 1.5 m. The length of the aluminium wire is 0.6 m. Therefore, the length of the steel wire (LS) can be calculated as:
\[
LS = 1.5 \, \text{m} - 0.6 \, \text{m} = 0.9 \, \text{m}
\]
### Step 2: Calculate the Linear Mass Densities
The linear mass density (μ) of a wire is given by the formula:
\[
\mu = \frac{\text{mass}}{\text{length}} = \frac{\rho \cdot A}{L}
\]
where:
- \( \rho \) is the density of the material,
- \( A \) is the cross-sectional area,
- \( L \) is the length of the wire.
For aluminium:
- Density \( \rho_A = 2.6 \times 10^3 \, \text{kg/m}^3 \)
- Cross-sectional area \( A = 10^{-6} \, \text{m}^2 \)
- Length \( L_A = 0.6 \, \text{m} \)
Calculating the linear mass density for aluminium:
\[
\mu_A = \frac{2.6 \times 10^3 \, \text{kg/m}^3 \times 10^{-6} \, \text{m}^2}{0.6 \, \text{m}} = \frac{2.6 \times 10^{-3}}{0.6} \approx 4.33 \times 10^{-3} \, \text{kg/m}
\]
For steel:
- Density \( \rho_S = 1.04 \times 10^4 \, \text{kg/m}^3 \)
- Length \( L_S = 0.9 \, \text{m} \)
Calculating the linear mass density for steel:
\[
\mu_S = \frac{1.04 \times 10^4 \, \text{kg/m}^3 \times 10^{-6} \, \text{m}^2}{0.9 \, \text{m}} = \frac{1.04 \times 10^{-2}}{0.9} \approx 1.16 \times 10^{-2} \, \text{kg/m}
\]
### Step 3: Establish the Relationship Between Frequencies
The frequency of the waves in each wire can be expressed as:
\[
f_A = \frac{n_A v_A}{2L_A} \quad \text{and} \quad f_S = \frac{n_S v_S}{2L_S}
\]
where \( n_A \) and \( n_S \) are the number of loops in aluminium and steel respectively, and \( v_A \) and \( v_S \) are the wave velocities in aluminium and steel.
Since both wires vibrate at the same frequency:
\[
f_A = f_S
\]
### Step 4: Calculate the Wave Velocities
The wave velocity in a wire is given by:
\[
v = \sqrt{\frac{T}{\mu}}
\]
where \( T \) is the tension in the wire. The tension due to the weight of 10 kg is:
\[
T = mg = 10 \, \text{kg} \times 10 \, \text{m/s}^2 = 100 \, \text{N}
\]
Calculating the velocities:
\[
v_A = \sqrt{\frac{100 \, \text{N}}{4.33 \times 10^{-3} \, \text{kg/m}}} \approx 148.5 \, \text{m/s}
\]
\[
v_S = \sqrt{\frac{100 \, \text{N}}{1.16 \times 10^{-2} \, \text{kg/m}}} \approx 91.5 \, \text{m/s}
\]
### Step 5: Set Up the Equation for Frequencies
From the frequency equations:
\[
\frac{n_A v_A}{2L_A} = \frac{n_S v_S}{2L_S}
\]
This simplifies to:
\[
\frac{n_A}{n_S} = \frac{v_S L_A}{v_A L_S}
\]
### Step 6: Substitute Values
Substituting the known values:
\[
\frac{n_A}{n_S} = \frac{91.5 \times 0.6}{148.5 \times 0.9} \approx \frac{54.9}{133.65} \approx 0.41
\]
This ratio suggests that the lowest integer values for \( n_A \) and \( n_S \) that satisfy this ratio are \( n_A = 1 \) and \( n_S = 3 \).
### Step 7: Calculate the Minimum Frequency
Using the frequency formula for aluminium:
\[
f_{min} = \frac{n_A v_A}{2L_A} = \frac{1 \times 148.5}{2 \times 0.6} \approx 123.75 \, \text{Hz}
\]
### Step 8: Determine the Total Number of Nodes
The total number of nodes in the system can be calculated as:
- For \( n_A = 1 \): 2 nodes (1 at each end)
- For \( n_S = 3 \): 4 nodes (3 loops create 4 nodes)
Thus, the total number of nodes:
\[
\text{Total Nodes} = 2 + 4 - 1 = 5
\]
### Final Answers
- The lowest frequency of excitation is approximately **123.75 Hz**.
- The total number of nodes at this frequency is **5**.
