When two sound waves with a phase difference of ` pi //2`, and each having amplitude A and frequency `omega` , are superimposed on each other, then the maximum amplitude and frequency of resultant wave is

Answer

To solve the problem of finding the maximum amplitude and frequency of the resultant wave when two sound waves with a phase difference of \( \frac{\pi}{2} \) are superimposed, we can follow these steps:

### Step 1: Understand the given parameters
We have two sound waves with:
- Amplitude \( A \)
- Frequency \( \omega \)
- Phase difference \( \phi = \frac{\pi}{2} \)

### Step 2: Write the formula for resultant amplitude
The resultant amplitude \( R \) when two waves are superimposed can be calculated using the formula:
\[ R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi)} \]
where \( A_1 \) and \( A_2 \) are the amplitudes of the two waves, and \( \phi \) is the phase difference.

### Step 3: Substitute the values into the formula
Since both waves have the same amplitude \( A \), we can set \( A_1 = A \) and \( A_2 = A \). Thus, the formula becomes:
\[ R = \sqrt{A^2 + A^2 + 2 A A \cos\left(\frac{\pi}{2}\right)} \]

### Step 4: Calculate the cosine of the phase difference
We know that:
\[ \cos\left(\frac{\pi}{2}\right) = 0 \]
Substituting this into our equation gives:
\[ R = \sqrt{A^2 + A^2 + 2 A^2 \cdot 0} \]

### Step 5: Simplify the expression
This simplifies to:
\[ R = \sqrt{A^2 + A^2} = \sqrt{2A^2} = \sqrt{2} A \]

### Step 6: Determine the frequency of the resultant wave
The frequency of the resultant wave remains the same as the frequency of the individual waves. Therefore, the resultant frequency is:
\[ \omega \]

### Final Result
The maximum amplitude of the resultant wave is \( R = \sqrt{2} A \) and the frequency is \( \omega \).

### Summary
- Maximum Amplitude: \( \sqrt{2} A \)
- Frequency: \( \omega \)

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