Two waves, travelling in the same direction through the same region, have equal frequencies, wavelengths and amplitudes. If the amplitude of each wave is 4 mm and the phase difference between the waves is `90^@`, what is the resultant amplitude ?

Explanation

To find the resultant amplitude of two waves with equal amplitudes, frequencies, and a phase difference of \(90^\circ\), we can use the formula for the resultant amplitude when two waves superimpose:

\[ A_r = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos \phi} \]

Where:
- \(A_r\) is the resultant amplitude,
- \(A_1\) and \(A_2\) are the amplitudes of the two waves,
- \(\phi\) is the phase difference between the two waves.

### Step-by-Step Solution:

**Step 1: Identify the given values.**
- Amplitude of each wave, \(A_1 = A_2 = 4 \, \text{mm}\)
- Phase difference, \(\phi = 90^\circ\)

**Step 2: Substitute the values into the formula.**
Since both amplitudes are equal, we can substitute \(A_1\) and \(A_2\) into the formula:

\[ A_r = \sqrt{4^2 + 4^2 + 2 \cdot 4 \cdot 4 \cdot \cos(90^\circ)} \]

**Step 3: Calculate the cosine of the phase difference.**
We know that:

\[ \cos(90^\circ) = 0 \]

**Step 4: Simplify the equation.**
Substituting \(\cos(90^\circ) = 0\) into the equation gives:

\[ A_r = \sqrt{4^2 + 4^2 + 2 \cdot 4 \cdot 4 \cdot 0} \]
\[ A_r = \sqrt{16 + 16 + 0} \]
\[ A_r = \sqrt{32} \]

**Step 5: Calculate the square root.**
\[ A_r = \sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2} \]

**Step 6: Calculate the numerical value.**
Using the approximate value of \(\sqrt{2} \approx 1.414\):

\[ A_r \approx 4 \cdot 1.414 \approx 5.656 \, \text{mm} \]

### Final Answer:
The resultant amplitude is approximately \(5.65 \, \text{mm}\).

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