A wave is represented by
`y_1 = 10 cos (5x + 25 t)`
where, x is measured in meters and t in seconds. A second wave for which
`y_2 = 20 cos (5x + 25t + pi/3)`
interferes with the first wave. Deduce the amplitude and phase of the resultant wave.

Explanation

To solve the problem of finding the amplitude and phase of the resultant wave formed by the superposition of two waves, we can follow these steps:

### Step 1: Identify the Amplitudes and Phases
The first wave is given by:
\[ y_1 = 10 \cos(5x + 25t) \]
This means the amplitude \( A_1 = 10 \) and the phase \( \phi_1 = 0 \).

The second wave is given by:
\[ y_2 = 20 \cos(5x + 25t + \frac{\pi}{3}) \]
This means the amplitude \( A_2 = 20 \) and the phase \( \phi_2 = \frac{\pi}{3} \).

### Step 2: Calculate the Resultant Amplitude
The resultant amplitude \( A_R \) of two waves can be calculated using the formula:
\[ A_R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi_2 - \phi_1)} \]
Substituting the values:
\[ A_R = \sqrt{10^2 + 20^2 + 2 \cdot 10 \cdot 20 \cdot \cos\left(\frac{\pi}{3}\right)} \]
Calculating each term:
- \( A_1^2 = 100 \)
- \( A_2^2 = 400 \)
- \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \)
- \( 2 \cdot 10 \cdot 20 \cdot \frac{1}{2} = 200 \)

Now substituting back into the equation:
\[ A_R = \sqrt{100 + 400 + 200} = \sqrt{700} \]

### Step 3: Calculate the Resultant Phase
To find the phase \( \alpha \) of the resultant wave, we can use the following formula:
\[ \tan(\alpha) = \frac{A_2 \sin(\phi_2)}{A_1 + A_2 \cos(\phi_2)} \]
Substituting the values:
\[ \tan(\alpha) = \frac{20 \sin\left(\frac{\pi}{3}\right)}{10 + 20 \cos\left(\frac{\pi}{3}\right)} \]
Calculating \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \) and \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ \tan(\alpha) = \frac{20 \cdot \frac{\sqrt{3}}{2}}{10 + 20 \cdot \frac{1}{2}} = \frac{10\sqrt{3}}{10 + 10} = \frac{10\sqrt{3}}{20} = \frac{\sqrt{3}}{2} \]

### Step 4: Find the Angle
Now to find \( \alpha \):
\[ \alpha = \tan^{-1}\left(\frac{\sqrt{3}}{2}\right) \]

### Final Result
Thus, the resultant amplitude and phase of the wave are:
- Amplitude: \( A_R = 10\sqrt{7} \, \text{meters} \)
- Phase: \( \alpha = \tan^{-1}\left(\frac{\sqrt{3}}{2}\right) \)