The tension in a stretched string fixed at both ends is changed by `2%`, the fundamental frequency is found to get changed by `15Hz`.
`(a)` wavelength of the string of fundamental frequency does not change
`(b)` velocity of propagation of wave changes by `2%`
`(c )` velocity of propagation of wave changes by `1%`
`(d)` original frequency is `1500Hz`

Options

A. `c` only correct
B. `c` and `d` are correct
C. `a,c` and `d` are correct
D. `b` and `d` are correct

Explanation

To solve the problem, we need to analyze the relationship between tension, frequency, and wave speed in a stretched string. Let's break it down step by step.

### Step 1: Understand the relationship between tension and frequency
The fundamental frequency \( f \) of a string fixed at both ends 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 mass per unit length of the string.

### Step 2: Analyze the change in tension
If the tension \( T \) is increased by \( 2\% \), we can express the new tension as:
\[ T' = 1.02T \]

### Step 3: Determine the change in frequency
The change in frequency due to the change in tension can be calculated using the formula for frequency:
\[ f' = \frac{1}{2L} \sqrt{\frac{T'}{\mu}} = \frac{1}{2L} \sqrt{\frac{1.02T}{\mu}} = \sqrt{1.02} \cdot f \]
The percentage change in frequency can be approximated as:
\[ \Delta f = f' - f = \left(\sqrt{1.02} - 1\right) f \]
Given that the change in frequency \( \Delta f \) is \( 15 \, \text{Hz} \), we can relate this to the original frequency.

### Step 4: Calculate the original frequency
Using the approximation \( \sqrt{1.02} \approx 1 + \frac{0.02}{2} = 1.01 \):
\[ \Delta f \approx 0.01 f \]
Setting this equal to \( 15 \, \text{Hz} \):
\[ 0.01 f = 15 \implies f = 1500 \, \text{Hz} \]

### Step 5: Analyze the change in wave speed
The wave speed \( v \) in the string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
When the tension increases by \( 2\% \):
\[ v' = \sqrt{\frac{T'}{\mu}} = \sqrt{\frac{1.02T}{\mu}} = \sqrt{1.02} \cdot v \]
The percentage change in wave speed is also approximately \( 1\% \).

### Conclusion
Based on the analysis:
- (a) The wavelength does not change, as the length of the string remains constant.
- (b) The velocity of propagation changes by \( 2\% \) (incorrect).
- (c) The velocity of propagation changes by \( 1\% \) (correct).
- (d) The original frequency is \( 1500 \, \text{Hz} \) (correct).

Thus, the correct answers are (a), (c), and (d).