The frequency of a tuning fork A is 5% greater than that of a standard fork K. The frequency of another fork B is 3% less than that of K. When A and B are vibrated simultaneously 4 beats per second are heard. Find the frequencies of A and B.

Options

A. 52.5 Hz, 48.5 Hz
B. 63.5 Hz, 79.5 Hz
C. 10.5 Hz, 101 Hz
D. 124 Hz, 120 Hz

Answer

To solve the problem, we need to find the frequencies of tuning forks A and B based on the information given about their relationship to a standard fork K and the number of beats produced when they are sounded together.

### Step-by-Step Solution:

1. **Define the Frequencies**:
- Let the frequency of the standard fork K be \( N_K \).
- The frequency of tuning fork A, which is 5% greater than K, can be expressed as:
\[ N_A = N_K + 0.05 N_K = 1.05 N_K = \frac{105}{100} N_K \]

2. **Calculate the Frequency of Fork B**:
- The frequency of tuning fork B, which is 3% less than K, can be expressed as:
\[ N_B = N_K - 0.03 N_K = 0.97 N_K = \frac{97}{100} N_K \]

3. **Use the Beats Formula**:
- The number of beats per second produced when two tuning forks are sounded together is given by the absolute difference of their frequencies:
\[ \text{Beats} = |N_A - N_B| = 4 \text{ beats/second} \]

4. **Set Up the Equation**:
- Substitute the expressions for \( N_A \) and \( N_B \) into the beats equation:
\[ |N_A - N_B| = \left| \frac{105}{100} N_K - \frac{97}{100} N_K \right| = 4 \]
- Simplifying the left side:
\[ \left| \frac{105 - 97}{100} N_K \right| = 4 \]
\[ \left| \frac{8}{100} N_K \right| = 4 \]

5. **Solve for \( N_K \)**:
- Multiply both sides by 100 to eliminate the fraction:
\[ 8 N_K = 400 \]
- Now, divide by 8:
\[ N_K = \frac{400}{8} = 50 \text{ Hz} \]

6. **Calculate Frequencies of A and B**:
- Now that we have \( N_K \), we can find \( N_A \) and \( N_B \):
\[ N_A = 1.05 \times 50 = 52.5 \text{ Hz} \]
\[ N_B = 0.97 \times 50 = 48.5 \text{ Hz} \]

### Final Answer:
- The frequency of tuning fork A is **52.5 Hz**.
- The frequency of tuning fork B is **48.5 Hz**.