Two notes A and B sounded together produce 2 beats per second. When notes B and C are sounded together 3 beats with per second are produced. The notes A and C separately produce the same number of beats with a standard tuning fork of frequnecy 456 Hz. the possible frequency of note B is

Answer

To solve the problem step by step, we will denote the frequencies of notes A, B, and C as \( F_A \), \( F_B \), and \( F_C \) respectively. We know the following from the problem statement:

1. The beat frequency between notes A and B is 2 beats per second.
2. The beat frequency between notes B and C is 3 beats per second.
3. Notes A and C produce the same number of beats with a standard tuning fork of frequency 456 Hz.

### Step 1: Set up the equations based on the beat frequencies

From the information given, we can express the relationships as follows:

- For notes A and B:
\[ |F_A - F_B| = 2 \quad \text{(1)} \]

- For notes B and C:
\[ |F_B - F_C| = 3 \quad \text{(2)} \]

### Step 2: Express \( F_A \) and \( F_C \) in terms of \( F_B \)

From equation (1), we have two cases:
1. \( F_A = F_B + 2 \)
2. \( F_A = F_B - 2 \)

From equation (2), we also have two cases:
1. \( F_C = F_B + 3 \)
2. \( F_C = F_B - 3 \)

### Step 3: Use the standard tuning fork frequency

The problem states that notes A and C separately produce the same number of beats with a standard tuning fork of frequency 456 Hz. This means:
\[ |F_A - 456| = |F_C - 456| \quad \text{(3)} \]

### Step 4: Substitute the expressions into equation (3)

We will consider each case for \( F_A \) and \( F_C \):

#### Case 1: \( F_A = F_B + 2 \) and \( F_C = F_B + 3 \)
Substituting into equation (3):
\[ |F_B + 2 - 456| = |F_B + 3 - 456| \]

This simplifies to:
\[ |F_B - 454| = |F_B - 453| \]

This leads to two scenarios:
1. \( F_B - 454 = F_B - 453 \) (not possible)
2. \( F_B - 454 = -(F_B - 453) \)
\[ 2F_B = 907 \implies F_B = 453.5 \text{ Hz} \]

#### Case 2: \( F_A = F_B + 2 \) and \( F_C = F_B - 3 \)
Substituting into equation (3):
\[ |F_B + 2 - 456| = |F_B - 3 - 456| \]
This simplifies to:
\[ |F_B - 454| = |F_B - 459| \]
This leads to two scenarios:
1. \( F_B - 454 = F_B - 459 \) (not possible)
2. \( F_B - 454 = -(F_B - 459) \)
\[ 2F_B = 913 \implies F_B = 456.5 \text{ Hz} \]

#### Case 3: \( F_A = F_B - 2 \) and \( F_C = F_B + 3 \)
Substituting into equation (3):
\[ |F_B - 2 - 456| = |F_B + 3 - 456| \]
This simplifies to:
\[ |F_B - 458| = |F_B - 453| \]
This leads to two scenarios:
1. \( F_B - 458 = F_B - 453 \) (not possible)
2. \( F_B - 458 = -(F_B - 453) \)
\[ 2F_B = 911 \implies F_B = 455.5 \text{ Hz} \]

#### Case 4: \( F_A = F_B - 2 \) and \( F_C = F_B - 3 \)
Substituting into equation (3):
\[ |F_B - 2 - 456| = |F_B - 3 - 456| \]
This simplifies to:
\[ |F_B - 458| = |F_B - 459| \]
This leads to two scenarios:
1. \( F_B - 458 = F_B - 459 \) (not possible)
2. \( F_B - 458 = -(F_B - 459) \)
\[ 2F_B = 917 \implies F_B = 458.5 \text{ Hz} \]

### Conclusion

The possible frequencies of note B are:
- \( 453.5 \text{ Hz} \)
- \( 456.5 \text{ Hz} \)
- \( 455.5 \text{ Hz} \)
- \( 458.5 \text{ Hz} \)