A tuning fork A sounded together with another fork B produces 5 beats/s. Fork A is in unison with 0.95 m length of a sonometer and fork B is in unison with 0.98 m length of the sonometer wire. Find the frequency of forks A and B.

Answer

To solve the problem, we need to find the frequencies of tuning forks A and B based on the information provided.

### Step 1: Understand the relationship between frequency and length
The frequency of a tuning fork is inversely proportional to the length of the sonometer wire when the tension and mass per unit length remain constant. This can be expressed as:
\[ f \propto \frac{1}{L} \]
or
\[ f \cdot L = k \]
where \( k \) is a constant.

### Step 2: Set up the equations for forks A and B
Given that fork A is in unison with a length of 0.95 m and fork B is in unison with a length of 0.98 m, we can write:
\[ f_A \cdot 0.95 = k \]
\[ f_B \cdot 0.98 = k \]

### Step 3: Equate the two equations
Since both expressions equal \( k \), we can set them equal to each other:
\[ f_A \cdot 0.95 = f_B \cdot 0.98 \]

### Step 4: Solve for \( f_A \) in terms of \( f_B \)
Rearranging the equation gives:
\[ f_A = f_B \cdot \frac{0.98}{0.95} \]

### Step 5: Use the beat frequency information
The problem states that the beat frequency is 5 beats per second. This means:
\[ |f_A - f_B| = 5 \]
We can express this as two equations:
1. \( f_A - f_B = 5 \)
2. \( f_B - f_A = 5 \)

### Step 6: Substitute \( f_A \) in the beat frequency equation
Using the first equation:
\[ f_A - f_B = 5 \]
Substituting \( f_A \) from Step 4:
\[ f_B \cdot \frac{0.98}{0.95} - f_B = 5 \]

### Step 7: Factor out \( f_B \)
This simplifies to:
\[ f_B \left( \frac{0.98}{0.95} - 1 \right) = 5 \]

### Step 8: Calculate the factor
Calculating the factor:
\[ \frac{0.98}{0.95} - 1 = \frac{0.98 - 0.95}{0.95} = \frac{0.03}{0.95} \]

### Step 9: Substitute back to find \( f_B \)
Now we have:
\[ f_B \cdot \frac{0.03}{0.95} = 5 \]
Solving for \( f_B \):
\[ f_B = 5 \cdot \frac{0.95}{0.03} \]
\[ f_B = 5 \cdot 31.67 \]
\[ f_B \approx 158.33 \, \text{Hz} \]

### Step 10: Find \( f_A \)
Now substitute \( f_B \) back into the equation for \( f_A \):
\[ f_A = f_B + 5 \]
\[ f_A = 158.33 + 5 \]
\[ f_A \approx 163.33 \, \text{Hz} \]

### Final Answer
- Frequency of fork A (\( f_A \)): **163.33 Hz**
- Frequency of fork B (\( f_B \)): **158.33 Hz**