In a stretched string under tension and fixed at both ends, the tension is increased by four times, and the length is doubled, the frequency

Options

A. Becomes half
B. Remains the same
C. Becomes twice
D. Becomes four times

Answer

To solve the problem, we need to analyze how the frequency of a stretched string changes when the tension and length are altered.

### Step-by-Step Solution:

1. **Understand the formula for frequency of a stretched string**:
The frequency \( F \) of a stretched string fixed at both ends is given by the formula:
\[ F = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( F \) = frequency
- \( L \) = length of the string
- \( T \) = tension in the string
- \( \mu \) = linear mass density of the string

2. **Identify the changes in tension and length**:
- The tension \( T \) is increased by 4 times, so the new tension \( T' = 4T \).
- The length \( L \) is doubled, so the new length \( L' = 2L \).

3. **Substitute the new values into the frequency formula**:
The new frequency \( F' \) can be expressed as:
\[ F' = \frac{1}{2L'} \sqrt{\frac{T'}{\mu}} = \frac{1}{2(2L)} \sqrt{\frac{4T}{\mu}} \]

4. **Simplify the expression**:
Substitute \( L' \) and \( T' \) into the equation:
\[ F' = \frac{1}{4L} \sqrt{\frac{4T}{\mu}} \]
This can be simplified further:
\[ F' = \frac{1}{4L} \cdot 2\sqrt{\frac{T}{\mu}} = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

5. **Relate the new frequency to the original frequency**:
Notice that:
\[ F' = F \]
This means that the new frequency \( F' \) is equal to the original frequency \( F \).

### Conclusion:
The frequency of the string does not change despite the increase in tension and the doubling of length. Therefore, the answer is that the frequency remains the same.