Two strings A and B made of same material are stretched by same tension. The radius of string A is double of the radius of B. A transverse wave travels on A with speed `v_A` and on B with speed `v_B`. The ratio `v_A/v_B` is

Options

A. `1/2`
B. 2
C. `1/4`
D. 4

Answer

To solve the problem, we need to find the ratio of the speeds of transverse waves on two strings A and B, given that they are made of the same material and stretched by the same tension, but have different radii.

### Step-by-step Solution:

1. **Understanding Wave Speed in Strings**:
The speed of a transverse wave on a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string and \( \mu \) is the linear mass density of the string.

2. **Linear Mass Density**:
The linear mass density \( \mu \) is defined as:
\[ \mu = \frac{m}{L} \]
where \( m \) is the mass of the string and \( L \) is its length. For a string of uniform cross-section, we can also express \( \mu \) in terms of the volume and the area of cross-section:
\[ \mu = \frac{\rho \cdot A}{L} \]
where \( \rho \) is the density of the material and \( A \) is the cross-sectional area of the string.

3. **Cross-sectional Area**:
The cross-sectional area \( A \) of a cylindrical string can be calculated using the radius \( r \):
\[ A = \pi r^2 \]

4. **Comparing the Two Strings**:
Let the radius of string B be \( r \). Then, the radius of string A is given as double that of string B:
\[ r_A = 2r \]
Therefore, the cross-sectional areas for the two strings are:
\[ A_A = \pi (2r)^2 = 4\pi r^2 \]
\[ A_B = \pi r^2 \]

5. **Calculating Linear Mass Densities**:
Now we can express the linear mass densities for both strings:
\[ \mu_A = \frac{\rho \cdot A_A}{L} = \frac{\rho \cdot 4\pi r^2}{L} \]
\[ \mu_B = \frac{\rho \cdot A_B}{L} = \frac{\rho \cdot \pi r^2}{L} \]

6. **Finding the Speeds**:
Since the tension \( T \) is the same for both strings, we can write the speeds as:
\[ v_A = \sqrt{\frac{T}{\mu_A}} = \sqrt{\frac{T}{\frac{\rho \cdot 4\pi r^2}{L}}} = \sqrt{\frac{TL}{\rho \cdot 4\pi r^2}} \]
\[ v_B = \sqrt{\frac{T}{\mu_B}} = \sqrt{\frac{T}{\frac{\rho \cdot \pi r^2}{L}}} = \sqrt{\frac{TL}{\rho \cdot \pi r^2}} \]

7. **Finding the Ratio of Speeds**:
Now, we can find the ratio \( \frac{v_A}{v_B} \):
\[ \frac{v_A}{v_B} = \frac{\sqrt{\frac{TL}{\rho \cdot 4\pi r^2}}}{\sqrt{\frac{TL}{\rho \cdot \pi r^2}}} = \frac{\sqrt{1/4}}{\sqrt{1}} = \frac{1}{2} \]

8. **Final Result**:
Therefore, the ratio of the speeds of the waves on strings A and B is:
\[ \frac{v_A}{v_B} = \frac{1}{2} \]