There are two wires of linear mass densities `lambda and 2lambda` and the tension in them are T and 2T, respectvely . Which wire is supposed to be the denser medium for transverse wave propagation ?

Options

A. `1^("st")` wire
B. `2^("nd")` wire
C. both are equally denser
D. nothing can be said in general

Answer

To determine which wire is the denser medium for transverse wave propagation, we will analyze the velocities of transverse waves in both wires using the formula for wave velocity.

### Step 1: Understand the formula for wave velocity
The velocity \( v \) of a transverse wave in a medium is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire and \( \mu \) is the linear mass density of the wire.

### Step 2: Calculate the velocity for the first wire
For the first wire:
- Linear mass density \( \mu_1 = \lambda \)
- Tension \( T_1 = T \)

Using the formula:
\[ v_1 = \sqrt{\frac{T_1}{\mu_1}} = \sqrt{\frac{T}{\lambda}} \]

### Step 3: Calculate the velocity for the second wire
For the second wire:
- Linear mass density \( \mu_2 = 2\lambda \)
- Tension \( T_2 = 2T \)

Using the formula:
\[ v_2 = \sqrt{\frac{T_2}{\mu_2}} = \sqrt{\frac{2T}{2\lambda}} = \sqrt{\frac{T}{\lambda}} \]

### Step 4: Compare the velocities
Now we have:
- \( v_1 = \sqrt{\frac{T}{\lambda}} \)
- \( v_2 = \sqrt{\frac{T}{\lambda}} \)

Both velocities are equal:
\[ v_1 = v_2 \]

### Step 5: Conclusion
Since the velocities of transverse waves in both wires are equal, we conclude that both wires allow transverse waves to propagate equally well. Therefore, both wires can be considered to be equally dense mediums for transverse wave propagation.

### Final Answer
Both wires are equally dense mediums for transverse wave propagation.

---