A steel wire of length `50sqrt(3) cm` is connected to an aluminium wire of length `60 cm` and stretched between two fixed supports. The tension produced is `104 N`, if the cross section area of each wire is `1mm^(2)`. If a transverse wave is set up in the wire, find the lowest frequency for which standing waves with node at the joint are produced . (density of aluminimum `= 2.6 gm//cm^(3)` and density of steel `= 7.8 gm//cm^(3)`).

Explanation

To solve the problem step by step, we need to find the lowest frequency for which standing waves with a node at the joint of the steel and aluminum wires are produced.

### Step 1: Understand the Setup
We have two wires:
- A steel wire of length \( L_1 = 50\sqrt{3} \, \text{cm} \)
- An aluminum wire of length \( L_2 = 60 \, \text{cm} \)

Both wires are under the same tension \( T = 104 \, \text{N} \) and have the same cross-sectional area \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \).

### Step 2: Calculate the Linear Densities
The linear density \( \mu \) of a wire is given by:
\[ \mu = \frac{\text{mass}}{\text{length}} = \frac{\rho \cdot A}{L} \]
Where \( \rho \) is the density of the material.

- For steel:
\[ \rho_{\text{steel}} = 7.8 \, \text{g/cm}^3 = 7800 \, \text{kg/m}^3 \]
\[ \mu_1 = \frac{7800 \times 1 \times 10^{-6}}{0.5\sqrt{3}} \, \text{kg/m} \]

- For aluminum:
\[ \rho_{\text{aluminum}} = 2.6 \, \text{g/cm}^3 = 2600 \, \text{kg/m}^3 \]
\[ \mu_2 = \frac{2600 \times 1 \times 10^{-6}}{0.6} \, \text{kg/m} \]

### Step 3: Calculate the Velocities of Waves in Each Wire
The velocity \( v \) of a wave in a wire is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]

- For steel:
\[ v_1 = \sqrt{\frac{104}{\mu_1}} \]

- For aluminum:
\[ v_2 = \sqrt{\frac{104}{\mu_2}} \]

### Step 4: Determine the Frequencies
The frequency \( f \) of the standing wave is given by:
\[ f = \frac{n v}{2L} \]
Where \( n \) is the mode number, \( v \) is the wave velocity, and \( L \) is the length of the wire.

For the lowest frequency with a node at the joint, we have:
\[ \frac{n_1 v_1}{2L_1} = \frac{n_2 v_2}{2L_2} \]

### Step 5: Relate the Frequencies
From the previous relation, we can derive:
\[ \frac{n_1}{n_2} = \frac{v_2 L_1}{v_1 L_2} \]

### Step 6: Substitute Values
Substituting the expressions for \( v_1 \) and \( v_2 \):
\[ \frac{n_1}{n_2} = \frac{\sqrt{\frac{104}{\mu_2}} \cdot L_1}{\sqrt{\frac{104}{\mu_1}} \cdot L_2} \]
This simplifies to:
\[ \frac{n_1}{n_2} = \frac{L_1}{L_2} \cdot \sqrt{\frac{\mu_1}{\mu_2}} \]

### Step 7: Calculate \( \frac{n_1}{n_2} \)
Substituting the lengths and densities:
\[ \frac{n_1}{n_2} = \frac{50\sqrt{3}}{60} \cdot \sqrt{\frac{7800}{2600}} = \frac{5\sqrt{3}}{6} \cdot \sqrt{3} = \frac{5}{2} \]

### Step 8: Choose the Lowest Values for \( n_1 \) and \( n_2 \)
To find the lowest frequency, we can take \( n_2 = 2 \) (the smallest integer) and \( n_1 = 5 \).

### Step 9: Calculate the Frequency
Using \( n_2 = 2 \):
\[ f = \frac{2 \cdot v_2}{2L_2} = \frac{v_2}{L_2} \]
Substituting \( v_2 \):
\[ f = \frac{1}{60} \sqrt{\frac{104}{\mu_2}} \]

### Step 10: Final Calculation
Substituting the values and calculating gives:
\[ f = \frac{1}{60} \sqrt{\frac{104}{\frac{2600 \times 1 \times 10^{-6}}{0.6}}} \]
This results in:
\[ f \approx \frac{1000}{3} \, \text{Hz} \]

### Final Answer
The lowest frequency for which standing waves with a node at the joint are produced is approximately:
\[ f \approx 333.33 \, \text{Hz} \]