A tuning fork of frequency 256 produces 4 beats per second with another tuning fork B. When the prongs of B are loaded with 1 gm wt, the number of beats is 1 per second and when loaded with 2 gm wt, the number of beats becomes 2 per second. What is the frequency of B?

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the concept of beats
When two tuning forks of different frequencies are sounded together, they produce beats. The number of beats per second is equal to the absolute difference between their frequencies.

### Step 2: Set up the equations
Let the frequency of tuning fork B be \( f_B \). According to the problem:
1. When tuning fork A (256 Hz) and tuning fork B are sounded together, they produce 4 beats per second. Therefore:
\[ |f_B - 256| = 4 \]
This gives us two possible equations:
\[ f_B - 256 = 4 \quad \text{(1)} \]
\[ 256 - f_B = 4 \quad \text{(2)} \]

### Step 3: Solve the equations
From equation (1):
\[ f_B = 256 + 4 = 260 \text{ Hz} \]
From equation (2):
\[ f_B = 256 - 4 = 252 \text{ Hz} \]

### Step 4: Analyze the effect of loading
The problem states that when the prongs of fork B are loaded with 1 gm weight, the number of beats is 1 per second:
\[ |f_B' - 256| = 1 \]
Where \( f_B' \) is the new frequency of fork B after loading with 1 gm. This gives:
\[ f_B' = 256 + 1 = 257 \text{ Hz} \quad \text{or} \quad f_B' = 256 - 1 = 255 \text{ Hz} \]

When loaded with 2 gm weight, the number of beats becomes 2 per second:
\[ |f_B'' - 256| = 2 \]
Where \( f_B'' \) is the frequency after loading with 2 gm. This gives:
\[ f_B'' = 256 + 2 = 258 \text{ Hz} \quad \text{or} \quad f_B'' = 256 - 2 = 254 \text{ Hz} \]

### Step 5: Determine the frequency of B
Since the frequency of tuning fork B increases with loading, we can conclude that:
- For 1 gm loading, \( f_B' \) must be greater than 256 Hz, so \( f_B' = 257 \text{ Hz} \).
- For 2 gm loading, \( f_B'' \) must also be greater than 256 Hz, so \( f_B'' = 258 \text{ Hz} \).

Since the frequency of B must be greater than 256 Hz, and we found that \( f_B = 260 \text{ Hz} \) from our first calculations, we can conclude that the frequency of tuning fork B is:
\[ \text{Frequency of B} = 260 \text{ Hz} \]

### Final Answer
The frequency of tuning fork B is **260 Hz**.

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