If the tension in a stretched string fixed at both ends is changed by `21%` , the fundamental frequency is found to increase by `15 Hz`, then the

Options

A. original frequency is 150 Hz
B. velocity of propagation of the transverse wave along the string increases by 5%.
C. velocity of propagation of the transverse wave along the string increases by 10% .
D. fundamental wavelength on the string does not change.

Explanation

To solve the problem step by step, we will analyze the relationship between tension, frequency, and wave velocity in a stretched string.

### Step 1: Understanding the relationship between tension and frequency
The fundamental frequency \( f \) of a string fixed at both ends is given by the formula:
\[ f = \frac{V}{2L} \]
where \( V \) is the velocity of the wave on the string and \( L \) is the length of the string.

### Step 2: Expressing wave velocity in terms of tension
The velocity \( V \) of transverse waves on the string can be expressed as:
\[ V = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string and \( \mu \) is the linear mass density of the string.

### Step 3: Analyzing the change in tension
If the tension in the string is increased by \( 21\% \), the new tension \( T_2 \) can be expressed as:
\[ T_2 = T_1 + 0.21 T_1 = 1.21 T_1 \]

### Step 4: Finding the new velocity
The new velocity \( V_2 \) can be calculated as:
\[ V_2 = \sqrt{\frac{T_2}{\mu}} = \sqrt{\frac{1.21 T_1}{\mu}} = \sqrt{1.21} \cdot \sqrt{\frac{T_1}{\mu}} = \sqrt{1.21} \cdot V_1 \]
Thus,
\[ V_2 = \frac{11}{10} V_1 \]

### Step 5: Calculating the percentage change in velocity
The percentage change in velocity can be calculated as:
\[ \text{Percentage Change} = \frac{V_2 - V_1}{V_1} \times 100 = \frac{\frac{11}{10} V_1 - V_1}{V_1} \times 100 = \left(\frac{1}{10}\right) \times 100 = 10\% \]

### Step 6: Relating frequency change to tension change
Since frequency is directly proportional to the square root of tension, we can express the relationship between the initial frequency \( f_1 \) and the new frequency \( f_2 \):
\[ \frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{1.21} = \frac{11}{10} \]
This implies:
\[ f_2 = \frac{11}{10} f_1 \]

### Step 7: Setting up the equation for frequency increase
Given that the fundamental frequency increases by \( 15 \, \text{Hz} \), we can write:
\[ f_2 = f_1 + 15 \]
Substituting \( f_2 \) from the previous equation:
\[ \frac{11}{10} f_1 = f_1 + 15 \]

### Step 8: Solving for the original frequency
Rearranging gives:
\[ \frac{11}{10} f_1 - f_1 = 15 \]
\[ \frac{1}{10} f_1 = 15 \implies f_1 = 150 \, \text{Hz} \]

### Conclusion
The original frequency \( f_1 \) is \( 150 \, \text{Hz} \).