Two sound waves of equal intensity I generates beats. The maximum intensity of sound produced in beats will be
I
4I
2I
I/2

Options

A. I
B. 4I
C. 2I
D. I/2

Explanation

To solve the problem of finding the maximum intensity of sound produced by two sound waves of equal intensity \(I\) that generate beats, we can follow these steps:

### Step 1: Understand the formula for intensity in superposition
When two sound waves interfere, the intensity of the resultant wave can be calculated using the formula:
\[ I_{\text{net}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
where \(I_1\) and \(I_2\) are the intensities of the individual waves and \(\phi\) is the phase difference between them.

### Step 2: Substitute the values for equal intensities
Since both sound waves have the same intensity \(I\), we have:
\[ I_1 = I_2 = I \]
Substituting these values into the formula gives:
\[ I_{\text{net}} = I + I + 2\sqrt{I \cdot I} \cos(\phi) \]

### Step 3: Simplify the equation
This simplifies to:
\[ I_{\text{net}} = 2I + 2\sqrt{I^2} \cos(\phi) \]
Since \(\sqrt{I^2} = I\), we can rewrite it as:
\[ I_{\text{net}} = 2I + 2I \cos(\phi) \]
This can be factored to:
\[ I_{\text{net}} = 2I(1 + \cos(\phi)) \]

### Step 4: Determine the maximum intensity
To find the maximum intensity, we need to maximize the term \(1 + \cos(\phi)\). The maximum value of \(\cos(\phi)\) is \(1\), which occurs when the waves are in phase. Therefore:
\[ 1 + \cos(\phi) = 1 + 1 = 2 \]
Substituting this back into the equation gives:
\[ I_{\text{max}} = 2I \cdot 2 = 4I \]

### Conclusion
Thus, the maximum intensity of sound produced in beats is:
\[ \boxed{4I} \]