To solve the problem, we need to analyze the situation step by step.
### Step 1: Understand the concept of beats
When two sound waves of slightly different frequencies interfere, they produce a phenomenon known as beats. The number of beats per second is equal to the absolute difference between the two frequencies.
### Step 2: Set up the equations for the first scenario
Let the initial frequency of the vibrating string be \( f_1 \) and the frequency of the tuning fork be \( f_2 = 256 \, \text{Hz} \). According to the problem, the string produces 2 beats per second with the tuning fork:
\[
|f_1 - f_2| = 2
\]
This gives us two possible equations:
1. \( f_1 - 256 = 2 \) (when \( f_1 > f_2 \))
2. \( 256 - f_1 = 2 \) (when \( f_2 > f_1 \))
### Step 3: Solve the equations
From the first equation:
\[
f_1 = 256 + 2 = 258 \, \text{Hz}
\]
From the second equation:
\[
f_1 = 256 - 2 = 254 \, \text{Hz}
\]
So, the possible values for \( f_1 \) are \( 258 \, \text{Hz} \) and \( 254 \, \text{Hz} \).
### Step 4: Analyze the second scenario
The problem states that slightly increasing the tension in the string produces 3 beats per second. When the tension is increased, the frequency of the string increases. Let’s denote the new frequency of the string as \( f_3 \).
Since the tension is increased, we can express \( f_3 \) as:
\[
f_3 = f_1 + a
\]
where \( a \) is a small positive value.
Now, we know that:
\[
|f_3 - f_2| = 3
\]
Substituting \( f_2 = 256 \, \text{Hz} \):
\[
|f_3 - 256| = 3
\]
### Step 5: Set up the equations for the second scenario
This gives us two new equations:
1. \( f_3 - 256 = 3 \)
2. \( 256 - f_3 = 3 \)
### Step 6: Solve the new equations
From the first equation:
\[
f_3 = 256 + 3 = 259 \, \text{Hz}
\]
From the second equation:
\[
f_3 = 256 - 3 = 253 \, \text{Hz}
\]
### Step 7: Substitute back to find \( a \)
Now, we substitute \( f_3 \) back into the equation \( f_3 = f_1 + a \) for both cases of \( f_1 \).
1. If \( f_1 = 258 \, \text{Hz} \):
\[
259 = 258 + a \implies a = 1 \, \text{Hz}
\]
2. If \( f_1 = 254 \, \text{Hz} \):
\[
253 = 254 + a \implies a = -1 \, \text{Hz} \quad (\text{not valid since } a \text{ must be positive})
\]
### Conclusion
The only valid initial frequency for the string is:
\[
\boxed{258 \, \text{Hz}}
\]
