Transverse waves are generated in two uniform wires `A` and `B` of the same material by attaching their free ends to a vibrating source of frequency `200Hz`. The cross sectional area of A is half that of B while the tension on A is twice that on B. The ratio of the wavelengths of the transverse waves in A and B is

Options

A. `1 : sqrt(2)`
B. `sqrt(2) : 1`
C. `1 : 2`
D. `2 : 1`

Answer

To solve the problem, we need to find the ratio of the wavelengths of the transverse waves in two wires A and B. We are given the following information:

- Frequency (f) = 200 Hz (same for both wires)
- Cross-sectional area of wire A (A_A) = 1/2 * Cross-sectional area of wire B (A_B)
- Tension in wire A (T_A) = 2 * Tension in wire B (T_B)

### Step 1: Define the relationship for wave velocity
The velocity (v) of a transverse wave in a string is given by the formula:

\[ v = \sqrt{\frac{T}{\mu}} \]

where T is the tension and μ (mu) is the mass per unit length of the string.

### Step 2: Express mass per unit length
The mass per unit length (μ) can be expressed as:

\[ \mu = \frac{m}{L} = \frac{\rho \cdot A \cdot L}{L} = \rho \cdot A \]

where:
- ρ (rho) is the density of the material,
- A is the cross-sectional area,
- L is the length of the wire (same for both wires).

### Step 3: Write the velocity for both wires
For wire A:

\[ v_A = \sqrt{\frac{T_A}{\mu_A}} = \sqrt{\frac{T_A}{\rho \cdot A_A}} \]

For wire B:

\[ v_B = \sqrt{\frac{T_B}{\mu_B}} = \sqrt{\frac{T_B}{\rho \cdot A_B}} \]

### Step 4: Substitute the values of tension and area
We know that:

- \( A_A = \frac{1}{2} A_B \)
- \( T_A = 2 T_B \)

Substituting these into the equations for velocities:

For wire A:

\[ v_A = \sqrt{\frac{2T_B}{\rho \cdot \frac{1}{2} A_B}} = \sqrt{\frac{2T_B \cdot 2}{\rho \cdot A_B}} = \sqrt{\frac{4T_B}{\rho \cdot A_B}} \]

For wire B:

\[ v_B = \sqrt{\frac{T_B}{\rho \cdot A_B}} \]

### Step 5: Find the ratio of velocities
Now, we can find the ratio of the velocities:

\[ \frac{v_A}{v_B} = \frac{\sqrt{\frac{4T_B}{\rho \cdot A_B}}}{\sqrt{\frac{T_B}{\rho \cdot A_B}}} = \sqrt{\frac{4T_B}{T_B}} = \sqrt{4} = 2 \]

### Step 6: Relate velocity and wavelength
The relationship between velocity (v), frequency (f), and wavelength (λ) is given by:

\[ v = f \cdot \lambda \]

Thus, we can express the wavelengths for both wires:

For wire A:

\[ \lambda_A = \frac{v_A}{f} \]

For wire B:

\[ \lambda_B = \frac{v_B}{f} \]

### Step 7: Find the ratio of wavelengths
Now, we can find the ratio of the wavelengths:

\[ \frac{\lambda_A}{\lambda_B} = \frac{v_A/f}{v_B/f} = \frac{v_A}{v_B} = 2 \]

### Conclusion
The ratio of the wavelengths of the transverse waves in wires A and B is:

\[ \frac{\lambda_A}{\lambda_B} = 2:1 \]