Two sound waves of intensity 2 `W//m^(2)` and `3W//m^(2)` meet at a point to produce a resultant intensity ` 5W//m^(2)`. The phase difference between two waves is

Options

A. `pi`
B. `(pi)/4`
C. `(pi)/(2)`
D. Zero

Answer

To solve the problem, we need to find the phase difference between two sound waves given their intensities and the resultant intensity. Let's break it down step by step.

### Step 1: Understand the given data
We have:
- Intensity of the first wave, \( I_1 = 2 \, \text{W/m}^2 \)
- Intensity of the second wave, \( I_2 = 3 \, \text{W/m}^2 \)
- Resultant intensity, \( I = 5 \, \text{W/m}^2 \)

### Step 2: Use the formula for resultant intensity
The formula for the resultant intensity when two waves interfere is given by:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
where \( \phi \) is the phase difference between the two waves.

### Step 3: Substitute the known values into the formula
Substituting the values we have into the formula:
\[ 5 = 2 + 3 + 2\sqrt{2 \cdot 3} \cos(\phi) \]

### Step 4: Simplify the equation
First, simplify the left side:
\[ 5 = 5 + 2\sqrt{6} \cos(\phi) \]
Now, subtract 5 from both sides:
\[ 0 = 2\sqrt{6} \cos(\phi) \]

### Step 5: Solve for \( \cos(\phi) \)
From the equation \( 0 = 2\sqrt{6} \cos(\phi) \), we can conclude:
\[ \cos(\phi) = 0 \]

### Step 6: Determine the phase difference
The cosine of an angle is zero at:
\[ \phi = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
For the principal value, we take:
\[ \phi = \frac{\pi}{2} \]

### Final Answer
The phase difference between the two waves is:
\[ \phi = \frac{\pi}{2} \text{ radians} \]

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