Transverse waves of same frequency are generated in two steel wires A and B . The diameter of A is twice of B and the tension in A is half that in B. The ratio of velocities of wave in A and B is

Options

A. `1:3sqrt(2)`
B. `1:2sqrt(2)`
C. `1:2`
D. `sqrt(2):1`

Answer

To solve the problem, we need to find the ratio of the velocities of waves in two steel wires A and B, given their diameters and tensions.

### Step 1: Understand the relationship between wave velocity, tension, and mass per unit length.
The velocity \( v \) of a wave in a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire and \( \mu \) is the mass per unit length of the wire.

### Step 2: Relate mass per unit length to diameter and density.
The mass per unit length \( \mu \) can be expressed in terms of the density \( \rho \) and the diameter \( D \) of the wire:
\[ \mu = \rho \cdot A = \rho \cdot \left(\frac{\pi D^2}{4}\right) \]
where \( A \) is the cross-sectional area of the wire.

### Step 3: Set up the equations for wires A and B.
Let the diameter of wire B be \( D \). Then the diameter of wire A is \( 2D \). The radii are:
\[ r_A = \frac{D_A}{2} = D \quad \text{and} \quad r_B = \frac{D_B}{2} = \frac{D}{2} \]

The mass per unit length for wires A and B can be expressed as:
\[ \mu_A = \rho \cdot \left(\frac{\pi (2D)^2}{4}\right) = \rho \cdot \pi D^2 \]
\[ \mu_B = \rho \cdot \left(\frac{\pi D^2}{4}\right) = \frac{\rho \cdot \pi D^2}{4} \]

### Step 4: Substitute the tensions and mass per unit lengths into the velocity formula.
Given that the tension in wire A is half that in wire B:
\[ T_A = \frac{1}{2} T_B \]

Now we can express the velocities:
\[ v_A = \sqrt{\frac{T_A}{\mu_A}} = \sqrt{\frac{\frac{1}{2} T_B}{\rho \cdot \pi D^2}} \]
\[ v_B = \sqrt{\frac{T_B}{\mu_B}} = \sqrt{\frac{T_B}{\frac{\rho \cdot \pi D^2}{4}}} \]

### Step 5: Find the ratio of the velocities.
Now, we can find the ratio \( \frac{v_A}{v_B} \):
\[ \frac{v_A}{v_B} = \frac{\sqrt{\frac{\frac{1}{2} T_B}{\rho \cdot \pi D^2}}}{\sqrt{\frac{T_B}{\frac{\rho \cdot \pi D^2}{4}}}} \]
This simplifies to:
\[ \frac{v_A}{v_B} = \sqrt{\frac{\frac{1}{2}}{\frac{4}{1}}} = \sqrt{\frac{1}{2} \cdot \frac{1}{4}} = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}} \]

### Final Step: Express the ratio in a simpler form.
Thus, the ratio of velocities \( v_A : v_B \) is:
\[ v_A : v_B = 1 : 2\sqrt{2} \]

### Final Answer:
The ratio of velocities of waves in A and B is \( 1 : 2\sqrt{2} \).