A fork A has frequency `2%` more than the standard fork and B has a frequency `3%` less than the frequency of same standard frok. The forks A and B when sounded together produced 6 beats/s. The frequency of fork A is

Options

A. `116.4 Hz`
B. `120 Hz`
C. `122.4 Hz`
D. `238.8 Hz`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define the standard frequency
Let the frequency of the standard fork be \( f \).

### Step 2: Determine the frequencies of forks A and B
- Fork A has a frequency that is 2% more than the standard fork:
\[ f_A = f + 0.02f = 1.02f \]
- Fork B has a frequency that is 3% less than the standard fork:
\[ f_B = f - 0.03f = 0.97f \]

### Step 3: Calculate the beat frequency
When two sound waves of different frequencies are sounded together, the beat frequency is given by the absolute difference of their frequencies. According to the problem, the beat frequency is 6 beats/s:
\[ |f_A - f_B| = 6 \]

### Step 4: Substitute the expressions for \( f_A \) and \( f_B \)
Substituting the expressions we found for \( f_A \) and \( f_B \):
\[ |1.02f - 0.97f| = 6 \]
This simplifies to:
\[ |0.05f| = 6 \]

### Step 5: Solve for \( f \)
Since \( 0.05f = 6 \):
\[ f = \frac{6}{0.05} = 120 \text{ Hz} \]

### Step 6: Calculate the frequency of fork A
Now that we have the standard frequency \( f \), we can find the frequency of fork A:
\[ f_A = 1.02f = 1.02 \times 120 = 122.4 \text{ Hz} \]

### Final Answer
The frequency of fork A is \( 122.4 \text{ Hz} \).

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