A and B are two wires whose fundamental frequencies are 256 and 382 Hz respectively. How many beats in 2 seconds will be heard by the third harmonic of A and second harmonic of B?

Options

A. 4
B. 8
C. 16
D. zero

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the fundamental frequencies
We are given the fundamental frequencies of wires A and B:
- Fundamental frequency of wire A, \( f_{0A} = 256 \, \text{Hz} \)
- Fundamental frequency of wire B, \( f_{0B} = 382 \, \text{Hz} \)

### Step 2: Calculate the third harmonic of wire A
The frequency of the third harmonic of wire A can be calculated using the formula:
\[ f_{3A} = 3 \times f_{0A} \]
Substituting the value of \( f_{0A} \):
\[ f_{3A} = 3 \times 256 = 768 \, \text{Hz} \]

### Step 3: Calculate the second harmonic of wire B
The frequency of the second harmonic of wire B can be calculated using the formula:
\[ f_{2B} = 2 \times f_{0B} \]
Substituting the value of \( f_{0B} \):
\[ f_{2B} = 2 \times 382 = 764 \, \text{Hz} \]

### Step 4: Calculate the beat frequency
The beat frequency is the absolute difference between the frequencies of the two harmonics:
\[ f_{\text{beat}} = |f_{3A} - f_{2B}| \]
Substituting the values calculated:
\[ f_{\text{beat}} = |768 - 764| = 4 \, \text{Hz} \]

### Step 5: Calculate the number of beats in 2 seconds
To find the total number of beats in 2 seconds, we multiply the beat frequency by the time duration:
\[ \text{Number of beats in 2 seconds} = f_{\text{beat}} \times 2 \]
Substituting the value of \( f_{\text{beat}} \):
\[ \text{Number of beats in 2 seconds} = 4 \times 2 = 8 \]

### Final Answer
The number of beats heard in 2 seconds is **8**.

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