If the length of a stretched string is shortened by `40 %` and the tension is increased by `44 %`, then the ratio of the final and initial fundamental frequencies is

Options

A. `3 : 4`
B. `4 : 3`
C. `1 : 3`
D. `2 :1`

Explanation

To find the ratio of the final and initial fundamental frequencies of a stretched string when its length is shortened by 40% and the tension is increased by 44%, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Fundamental Frequency**:
The fundamental frequency \( f \) of a stretched string is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( L \) is the length of the string, \( T \) is the tension, and \( \mu \) is the linear mass density.

2. **Define Initial and Final Conditions**:
Let:
- Initial length \( L_1 = L \)
- Final length \( L_2 = L_1 - 0.4L_1 = 0.6L_1 = 0.6L \)
- Initial tension \( T_1 = T \)
- Final tension \( T_2 = T_1 + 0.44T_1 = 1.44T_1 = 1.44T \)

3. **Set Up the Ratio of Frequencies**:
We want to find the ratio of the final frequency \( f_2 \) to the initial frequency \( f_1 \):
\[ \frac{f_2}{f_1} = \frac{\frac{1}{2L_2} \sqrt{\frac{T_2}{\mu}}}{\frac{1}{2L_1} \sqrt{\frac{T_1}{\mu}}} \]

4. **Simplify the Ratio**:
This simplifies to:
\[ \frac{f_2}{f_1} = \frac{L_1}{L_2} \cdot \sqrt{\frac{T_2}{T_1}} \]

5. **Substitute the Values**:
Substitute \( L_2 = 0.6L_1 \) and \( T_2 = 1.44T_1 \):
\[ \frac{f_2}{f_1} = \frac{L_1}{0.6L_1} \cdot \sqrt{\frac{1.44T_1}{T_1}} = \frac{1}{0.6} \cdot \sqrt{1.44} \]

6. **Calculate the Values**:
- \( \frac{1}{0.6} = \frac{10}{6} = \frac{5}{3} \)
- \( \sqrt{1.44} = 1.2 \)

Therefore:
\[ \frac{f_2}{f_1} = \frac{5}{3} \cdot 1.2 = \frac{5 \cdot 1.2}{3} = \frac{6}{3} = 2 \]

7. **Final Result**:
The ratio of the final and initial fundamental frequencies is:
\[ \frac{f_2}{f_1} = 2 \]

### Conclusion:
The ratio of the final and initial fundamental frequencies is \( 2:1 \).

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