The frequency of tuning fork A is 2% more than the frequency of a standard tuning fork. The frequency of a tuning fork B is 3% less than the frequency of the same standard tuning fork. If 6 beat/s are heard when the tuning fork A and B are excited , then frequency of A will be

Options

A. (a)120 Hz
B. (b)122.4 Hz
C. (c)116. 4 Hz
D. (d)130 Hz

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Define the standard frequency
Let the frequency of the standard tuning fork be denoted as \( n_s \).

### Step 2: Calculate the frequency of tuning fork A
Since tuning fork A is 2% more than the standard frequency, we can express its frequency as:
\[ n_A = n_s + 0.02 n_s = 1.02 n_s \]

### Step 3: Calculate the frequency of tuning fork B
Tuning fork B is 3% less than the standard frequency, so its frequency can be expressed as:
\[ n_B = n_s - 0.03 n_s = 0.97 n_s \]

### Step 4: Set up the equation for beat frequency
The beat frequency \( f_{beat} \) is given by the absolute difference between the frequencies of A and B:
\[ f_{beat} = |n_A - n_B| = 6 \text{ beats/s} \]
Substituting the expressions for \( n_A \) and \( n_B \):
\[ |1.02 n_s - 0.97 n_s| = 6 \]
This simplifies to:
\[ |0.05 n_s| = 6 \]

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

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

### Final Answer
The frequency of tuning fork A is:
\[ \boxed{122.4 \text{ Hz}} \]