Two strings A and B, made of the same material, have equal lengths. 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 velocities of transverse waves in A and B is

Options

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

Answer

To solve the problem, we need to find the ratio of the velocities of transverse waves in two strings A and B, given the conditions about their tension and cross-sectional areas.

### Step-by-Step Solution:

1. **Understanding the Formula for Wave Velocity**:
The velocity \( V \) of a transverse wave in a string is given by the formula:
\[ V = \sqrt{\frac{T}{A}} \]
where \( T \) is the tension in the string and \( A \) is the cross-sectional area.

2. **Setting Up the Ratios**:
We want to find the ratio of the velocities of the two strings:
\[ \frac{V_A}{V_B} = \sqrt{\frac{T_A}{A_A}} \div \sqrt{\frac{T_B}{A_B}} = \sqrt{\frac{T_A \cdot A_B}{T_B \cdot A_A}} \]

3. **Substituting the Given Values**:
- It is given that the tension in string A is twice that of string B:
\[ T_A = 2T_B \]
- The cross-sectional area of string A is half that of string B:
\[ A_A = \frac{1}{2} A_B \]

4. **Plugging in the Values**:
Now we substitute these values into the ratio:
\[ \frac{V_A}{V_B} = \sqrt{\frac{(2T_B) \cdot A_B}{T_B \cdot \left(\frac{1}{2} A_B\right)}} \]

5. **Simplifying the Expression**:
- The \( T_B \) in the numerator and denominator cancels out:
\[ \frac{V_A}{V_B} = \sqrt{\frac{2 \cdot A_B}{\frac{1}{2} A_B}} = \sqrt{\frac{2}{\frac{1}{2}}} = \sqrt{4} = 2 \]

6. **Final Ratio**:
Thus, the ratio of the velocities of transverse waves in strings A and B is:
\[ \frac{V_A}{V_B} = 2 \]
This can be expressed as:
\[ V_A : V_B = 2 : 1 \]

### Conclusion:
The ratio of the velocities of transverse waves in strings A and B is \( 2 : 1 \).

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