Four harmonic waves of equal freuencies and equal intensity `I_(0)` have phase angles `0, (pi)/(3), (2pi)/(3)` and `pi`. When they are superposed, the intensity of the resulting wave is `nI_(0)`. The value of `n` is

Explanation

To solve the problem, we need to find the value of \( n \) when four harmonic waves with equal frequencies and intensities are superposed. The phases of the waves are given as \( 0, \frac{\pi}{3}, \frac{2\pi}{3}, \) and \( \pi \).

### Step-by-Step Solution:

1. **Understanding the Relationship Between Intensity and Amplitude**:
The intensity \( I \) of a wave is proportional to the square of its amplitude \( A \):
\[ I \propto A^2 \]
Therefore, if the intensity of each wave is \( I_0 \), we can express this as:
\[ I_0 = k A_0^2 \]
where \( k \) is a proportionality constant.

2. **Expressing the Amplitudes of the Waves**:
Let each wave have an amplitude \( A_0 \). The four waves can be represented as:
- Wave 1: \( A_0 \) at phase \( 0 \)
- Wave 2: \( A_0 \) at phase \( \frac{\pi}{3} \)
- Wave 3: \( A_0 \) at phase \( \frac{2\pi}{3} \)
- Wave 4: \( A_0 \) at phase \( \pi \)

3. **Calculating the Resultant Amplitude**:
We can calculate the resultant amplitude by considering the x and y components of each wave. The waves can be represented in terms of their components:
- Wave 1: \( A_0 \) contributes \( (A_0, 0) \)
- Wave 2: \( A_0 \) contributes \( \left(A_0 \cos\frac{\pi}{3}, A_0 \sin\frac{\pi}{3}\right) \)
- Wave 3: \( A_0 \) contributes \( \left(A_0 \cos\frac{2\pi}{3}, A_0 \sin\frac{2\pi}{3}\right) \)
- Wave 4: \( A_0 \) contributes \( (-A_0, 0) \)

Now, calculating the x-components:
\[ A_{x} = A_0 + A_0 \cos\frac{\pi}{3} + A_0 \cos\frac{2\pi}{3} - A_0 \]
Since \( \cos\frac{\pi}{3} = \frac{1}{2} \) and \( \cos\frac{2\pi}{3} = -\frac{1}{2} \):
\[ A_{x} = A_0 + \frac{A_0}{2} - \frac{A_0}{2} - A_0 = 0 \]

Now, calculating the y-components:
\[ A_{y} = A_0 \sin\frac{\pi}{3} + A_0 \sin\frac{2\pi}{3} \]
Since \( \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} \) and \( \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2} \):
\[ A_{y} = A_0 \frac{\sqrt{3}}{2} + A_0 \frac{\sqrt{3}}{2} = A_0 \sqrt{3} \]

4. **Resultant Amplitude**:
The resultant amplitude \( A_R \) is given by:
\[ A_R = \sqrt{A_x^2 + A_y^2} = \sqrt{0^2 + (A_0 \sqrt{3})^2} = A_0 \sqrt{3} \]

5. **Calculating the Resultant Intensity**:
The intensity of the resultant wave \( I_R \) can be calculated as:
\[ I_R \propto A_R^2 = (A_0 \sqrt{3})^2 = 3 A_0^2 \]
Since \( I_0 = k A_0^2 \), we have:
\[ I_R = 3 k A_0^2 = 3 I_0 \]

6. **Finding the Value of \( n \)**:
From the above, we find that the intensity of the resultant wave is:
\[ I_R = n I_0 \implies n = 3 \]

### Final Answer:
The value of \( n \) is \( 3 \).