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 when the length of a stretched string is shortened by 40% and the tension is increased by 44%, we can follow these steps:

### Step 1: Understand the formula for fundamental frequency
The fundamental frequency \( N \) of a stretched string is given by the formula:
\[ N = \frac{1}{2L} \sqrt{\frac{T}{M}} \]
where:
- \( L \) is the length of the string,
- \( T \) is the tension in the string,
- \( M \) is the mass per unit length (which remains constant).

### Step 2: Define initial and final conditions
Let:
- \( L_1 \) = initial length of the string,
- \( T_1 \) = initial tension,
- \( N_1 \) = initial fundamental frequency,
- \( L_2 \) = final length of the string,
- \( T_2 \) = final tension,
- \( N_2 \) = final fundamental frequency.

### Step 3: Calculate the new length and tension
Given that the length is shortened by 40%, we can express the final length as:
\[ L_2 = L_1 - 0.4L_1 = 0.6L_1 \]

Given that the tension is increased by 44%, we can express the final tension as:
\[ T_2 = T_1 + 0.44T_1 = 1.44T_1 \]

### Step 4: Write the expressions for initial and final frequencies
Using the formula for frequency, we have:
\[ N_1 = \frac{1}{2L_1} \sqrt{\frac{T_1}{M}} \]
\[ N_2 = \frac{1}{2L_2} \sqrt{\frac{T_2}{M}} \]

### Step 5: Substitute \( L_2 \) and \( T_2 \) into the frequency formula
Substituting \( L_2 \) and \( T_2 \) into the equation for \( N_2 \):
\[ N_2 = \frac{1}{2(0.6L_1)} \sqrt{\frac{1.44T_1}{M}} \]

### Step 6: Simplify the expression for \( N_2 \)
\[ N_2 = \frac{1}{1.2L_1} \sqrt{\frac{1.44T_1}{M}} = \frac{1.2}{1.2L_1} \sqrt{\frac{T_1}{M}} = \frac{1.2}{1.2} \cdot N_1 \]
\[ N_2 = 2N_1 \]

### Step 7: Find the ratio of final to initial frequencies
The ratio of the final frequency to the initial frequency is:
\[ \frac{N_2}{N_1} = 2 \]

### Conclusion
Thus, the ratio of the final and initial fundamental frequencies is:
\[ \frac{N_2}{N_1} = 2:1 \]