Consider a wire with density (d) and stress `(sigma)` . For the same density . if the stress increases 2 times , the speed of the transverse waves along the wire change by

Options

A. `sqrt2`
B. `1/sqrt2`
C. 2
D. `1/2`

Answer

To solve the problem, we need to determine how the speed of transverse waves in a wire changes when the stress applied to the wire is doubled while keeping the density constant.

### Step-by-Step Solution:

1. **Understand the relationship between speed, stress, and density**:
The speed of transverse waves \( V \) in a wire is given by the formula:
\[ V = \sqrt{\frac{\sigma}{d}} \]
where \( \sigma \) is the stress and \( d \) is the density of the wire.

2. **Identify the initial conditions**:
Let the initial stress be \( \sigma_1 = \sigma \) and the initial speed be \( V_1 \):
\[ V_1 = \sqrt{\frac{\sigma}{d}} \]

3. **Change the stress**:
If the stress is doubled, then the new stress \( \sigma_2 \) becomes:
\[ \sigma_2 = 2\sigma \]

4. **Calculate the new speed**:
Substitute the new stress into the speed formula:
\[ V_2 = \sqrt{\frac{\sigma_2}{d}} = \sqrt{\frac{2\sigma}{d}} \]

5. **Relate the new speed to the initial speed**:
We can express \( V_2 \) in terms of \( V_1 \):
\[ V_2 = \sqrt{2} \cdot \sqrt{\frac{\sigma}{d}} = \sqrt{2} \cdot V_1 \]

6. **Conclusion**:
Since \( V_1 = \sqrt{\frac{\sigma}{d}} \), we find that:
\[ V_2 = \sqrt{2} \cdot V_1 \]
This means that when the stress is doubled, the speed of the transverse wave increases by a factor of \( \sqrt{2} \).

### Final Answer:
The speed of the transverse waves along the wire changes by a factor of \( \sqrt{2} \).