In a stretched wire under tension and fixed at both ends, the area of cross section of the wire is halved and the tension is doubled. The frequency of the wire will be

Options

A. twice
B. half
C. three tiems
D. 3 times

Explanation

To solve the problem, we need to analyze how the frequency of a stretched wire changes when the area of cross-section is halved and the tension is doubled.

### Step-by-Step Solution:

1. **Understand the Frequency Formula**:
The frequency \( F \) of a stretched wire fixed at both ends is given by the formula:
\[ F = \frac{1}{2L} \sqrt{\frac{T}{M}} \]
where:
- \( L \) = length of the wire
- \( T \) = tension in the wire
- \( M \) = mass per unit length of the wire

2. **Express Mass per Unit Length**:
The mass per unit length \( M \) can be expressed in terms of the cross-sectional area \( A \) and the density \( \rho \) of the material:
\[ M = \rho \cdot A \]

3. **Substitute for Mass in the Frequency Formula**:
Substitute \( M \) in the frequency formula:
\[ F = \frac{1}{2L} \sqrt{\frac{T}{\rho A}} \]

4. **Initial Conditions**:
Let the initial tension be \( T_1 = T \) and the initial area be \( A_1 = A \). The initial frequency \( F_1 \) can be written as:
\[ F_1 = \frac{1}{2L} \sqrt{\frac{T}{\rho A}} \]

5. **New Conditions**:
According to the problem:
- The area of cross-section is halved: \( A_2 = \frac{A}{2} \)
- The tension is doubled: \( T_2 = 2T \)

6. **Calculate the New Frequency**:
Substitute the new values into the frequency formula:
\[ F_2 = \frac{1}{2L} \sqrt{\frac{2T}{\rho \cdot \frac{A}{2}}} \]
Simplifying this:
\[ F_2 = \frac{1}{2L} \sqrt{\frac{2T \cdot 2}{\rho A}} = \frac{1}{2L} \sqrt{\frac{4T}{\rho A}} = \frac{2}{2L} \sqrt{\frac{T}{\rho A}} = 2F_1 \]

7. **Conclusion**:
The new frequency \( F_2 \) is twice the original frequency \( F_1 \):
\[ F_2 = 2F_1 \]

### Final Answer:
The frequency of the wire will be **twice the original frequency**.