The frequency of tuning forks A and B are respectively `3%` more and `2%` less than the frequency of tuning fork `C`. When A and B are simultaneously excited, 5 beats per second are produced. Then the frequency of the tuning fork `A` (in Hz)` Is

Options

A. 98
B. 100
C. 103
D. 105

Answer

To find the frequency of tuning fork A, we will follow these steps:

### Step 1: Define the frequencies of the tuning forks
Let the frequency of tuning fork C be \( f_C \). According to the problem:
- The frequency of tuning fork A, \( f_A \), is 3% more than \( f_C \):
\[ f_A = f_C + 0.03 f_C = 1.03 f_C \]
- The frequency of tuning fork B, \( f_B \), is 2% less than \( f_C \):
\[ f_B = f_C - 0.02 f_C = 0.98 f_C \]

### Step 2: Determine the beat frequency
When tuning forks A and B are excited together, they produce a beat frequency of 5 beats per second. The beat frequency is given by the absolute difference of their frequencies:
\[ |f_A - f_B| = 5 \text{ Hz} \]

### Step 3: Substitute the expressions for \( f_A \) and \( f_B \)
Substituting the expressions we found for \( f_A \) and \( f_B \):
\[ |1.03 f_C - 0.98 f_C| = 5 \]
This simplifies to:
\[ |0.05 f_C| = 5 \]

### Step 4: Solve for \( f_C \)
Since \( 0.05 f_C = 5 \), we can solve for \( f_C \):
\[ f_C = \frac{5}{0.05} = 100 \text{ Hz} \]

### Step 5: Calculate the frequency of tuning fork A
Now that we have \( f_C \), we can find \( f_A \):
\[ f_A = 1.03 f_C = 1.03 \times 100 = 103 \text{ Hz} \]

### Final Answer
The frequency of tuning fork A is \( \boxed{103} \text{ Hz} \).

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