If the tension in a string is increased by `21` percent, the fundamental frequency of the string changes by `15 Hz`. Which of the following statements will also be correct?

Options

A. The original fundamental frequency is nearly `150 Hz`
B. The velocity of propagation changes nearly by `4.5%`
C. The velocity of propagation changes nearly by `10%`
D. The fundamental wavelength changes nearly by `10%`

Explanation

To solve the problem, we need to analyze the relationship between tension, fundamental frequency, and wave velocity in a string. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between tension and frequency
The fundamental frequency \( f \) of a vibrating 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 in the string,
- \( \mu \) is the linear mass density of the string.

From this formula, we can see that the fundamental frequency \( f \) is directly proportional to the square root of the tension \( T \):
\[ f \propto \sqrt{T} \]

### Step 2: Calculate the new tension
If the tension in the string is increased by 21%, the new tension \( T' \) can be expressed as:
\[ T' = T + 0.21T = 1.21T \]

### Step 3: Relate the change in frequency to the change in tension
The change in frequency due to the increase in tension is given as 15 Hz. Therefore, the new frequency \( f' \) can be expressed as:
\[ f' = f + 15 \]

### Step 4: Express the new frequency in terms of the new tension
Using the relationship derived in Step 1, we can express the new frequency in terms of the new tension:
\[ f' = \frac{1}{2L} \sqrt{\frac{T'}{\mu}} = \frac{1}{2L} \sqrt{\frac{1.21T}{\mu}} = \sqrt{1.21} \cdot \frac{1}{2L} \sqrt{\frac{T}{\mu}} = \sqrt{1.21} \cdot f \]
Since \( \sqrt{1.21} = 1.1 \), we have:
\[ f' = 1.1f \]

### Step 5: Set up the equation
Now we can set up the equation using the expressions for \( f' \):
\[ f + 15 = 1.1f \]
Rearranging gives:
\[ 15 = 1.1f - f \]
\[ 15 = 0.1f \]
Thus, solving for \( f \):
\[ f = \frac{15}{0.1} = 150 \text{ Hz} \]

### Step 6: Calculate the change in velocity
The velocity \( v \) of the wave on the string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
The new velocity \( v' \) with the new tension \( T' \) is:
\[ v' = \sqrt{\frac{T'}{\mu}} = \sqrt{\frac{1.21T}{\mu}} = \sqrt{1.21} \cdot v = 1.1v \]
The percentage change in velocity is:
\[ \text{Percentage change} = \frac{v' - v}{v} \times 100 = \frac{1.1v - v}{v} \times 100 = \frac{0.1v}{v} \times 100 = 10\% \]

### Conclusion
From the analysis, we find:
- The original fundamental frequency \( f \) is 150 Hz (Option A is correct).
- The percentage change in wave velocity is 10% (Option C is correct).

Thus, the correct statements are:
- A: The original fundamental frequency is 150 Hz.
- C: The percentage change in velocity is 10%.