A tuning fork A produces 5 beats/sec with another tuning fork B of frequency 256 Hz. If tuning fork A produces 1 beats/sec with some other tuning fork C of frequency 250 Hz, then calculate the frequency of fork A.

Answer

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

### Step 1: Understand the beat frequency concept
The beat frequency is the absolute difference between the frequencies of two tuning forks. If tuning fork A produces 5 beats per second with tuning fork B (which has a frequency of 256 Hz), we can express this as:
\[ |f_A - f_B| = 5 \]
where \( f_B = 256 \, \text{Hz} \).

### Step 2: Set up the equations for tuning fork A and B
From the beat frequency condition, we can derive two possible equations for the frequency of tuning fork A:
1. \( f_A = f_B + 5 \)
2. \( f_A = f_B - 5 \)

Substituting \( f_B = 256 \, \text{Hz} \):
1. \( f_A = 256 + 5 = 261 \, \text{Hz} \)
2. \( f_A = 256 - 5 = 251 \, \text{Hz} \)

### Step 3: Analyze the second condition with tuning fork C
Tuning fork A produces 1 beat per second with tuning fork C, which has a frequency of 250 Hz. This gives us another equation:
\[ |f_A - f_C| = 1 \]
where \( f_C = 250 \, \text{Hz} \).

### Step 4: Set up the equations for tuning fork A and C
From the beat frequency condition with tuning fork C, we can derive two more equations:
1. \( f_A = f_C + 1 \)
2. \( f_A = f_C - 1 \)

Substituting \( f_C = 250 \, \text{Hz} \):
1. \( f_A = 250 + 1 = 251 \, \text{Hz} \)
2. \( f_A = 250 - 1 = 249 \, \text{Hz} \)

### Step 5: Find the common frequency
Now we have four possible frequencies for tuning fork A:
- From tuning fork B: \( 261 \, \text{Hz} \) or \( 251 \, \text{Hz} \)
- From tuning fork C: \( 251 \, \text{Hz} \) or \( 249 \, \text{Hz} \)

The only common frequency that satisfies both conditions is:
\[ f_A = 251 \, \text{Hz} \]

### Final Answer
The frequency of tuning fork A is \( 251 \, \text{Hz} \).

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