Tuning fork A and B produce 4 beats when sounded together. When they are tuned with saem sonometer wire, the resonance length are 128 cm and 130 cm respectively. The frequency of A is :

Options

A. 240 Hz
B. 250 Hz
C. 260 Hz
D. 270 Hz

Answer

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

### Step-by-Step Solution:

1. **Understand the Beat Frequency**:
The beat frequency is the difference in frequencies of two tuning forks. Given that the beat frequency is 4 Hz, we can express this relationship mathematically:
\[ |f_B - f_A| = 4 \]
where \(f_A\) is the frequency of tuning fork A and \(f_B\) is the frequency of tuning fork B.

2. **Use the Sonometer Formula**:
The frequency of a tuning fork can be calculated using the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \(L\) is the resonance length,
- \(T\) is the tension in the wire,
- \(\mu\) is the linear mass density of the wire.

3. **Calculate Frequencies for A and B**:
For tuning fork A (resonance length \(L_A = 128 \, \text{cm} = 1.28 \, \text{m}\)):
\[ f_A = \frac{1}{2 \times 1.28} \sqrt{\frac{T}{\mu}} \]
For tuning fork B (resonance length \(L_B = 130 \, \text{cm} = 1.30 \, \text{m}\)):
\[ f_B = \frac{1}{2 \times 1.30} \sqrt{\frac{T}{\mu}} \]

4. **Express Frequencies in Terms of Tension and Mass Density**:
Let \(k = \sqrt{\frac{T}{\mu}}\), then:
\[ f_A = \frac{k}{2 \times 1.28} \]
\[ f_B = \frac{k}{2 \times 1.30} \]

5. **Set Up the Beat Frequency Equation**:
Using the beat frequency equation:
\[ \left| \frac{k}{2 \times 1.30} - \frac{k}{2 \times 1.28} \right| = 4 \]
Simplifying this gives:
\[ \frac{k}{2} \left( \frac{1}{1.30} - \frac{1}{1.28} \right) = 4 \]

6. **Calculate the Difference**:
Calculate the difference:
\[ \frac{1}{1.30} - \frac{1}{1.28} = \frac{1.28 - 1.30}{1.30 \times 1.28} = \frac{-0.02}{1.664} \approx -0.012 \]
Thus,
\[ \frac{k}{2} \times (-0.012) = 4 \]
Rearranging gives:
\[ k = \frac{4 \times 2}{-0.012} \approx -666.67 \text{ (taking absolute value)} \]

7. **Calculate Frequency of A**:
Now substituting back to find \(f_A\):
\[ f_A = \frac{k}{2 \times 1.28} = \frac{666.67}{2 \times 1.28} \approx 260 \text{ Hz} \]

### Final Answer:
The frequency of tuning fork A is approximately **260 Hz**.