Two waves represented by `y_(i)=3sin(200x-150t) and y_(2)=3cos(200x-150t)` are superposed where x and y are in metre and t is in second. Calculate the amplitude of resultant wave

Explanation

To find the amplitude of the resultant wave formed by the superposition of the two waves given by \( y_1 = 3\sin(200x - 150t) \) and \( y_2 = 3\cos(200x - 150t) \), we can follow these steps:

### Step 1: Convert the cosine wave to sine form
The wave \( y_2 \) is given in cosine form. We can convert it to sine form using the identity:
\[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \]
Thus, we can rewrite \( y_2 \) as:
\[ y_2 = 3\cos(200x - 150t) = 3\sin\left(200x - 150t + \frac{\pi}{2}\right) \]

### Step 2: Identify the amplitudes and phase angles
From the equations:
- For \( y_1 = 3\sin(200x - 150t) \):
- Amplitude \( a_1 = 3 \)
- Phase angle \( \phi_1 = 0 \)

- For \( y_2 = 3\sin\left(200x - 150t + \frac{\pi}{2}\right) \):
- Amplitude \( a_2 = 3 \)
- Phase angle \( \phi_2 = \frac{\pi}{2} \)

### Step 3: Calculate the phase difference
The phase difference \( \Delta \phi \) between the two waves is:
\[ \Delta \phi = \phi_2 - \phi_1 = \frac{\pi}{2} - 0 = \frac{\pi}{2} \]

### Step 4: Use the formula for resultant amplitude
The resultant amplitude \( A \) of two superposed waves can be calculated using the formula:
\[ A = \sqrt{a_1^2 + a_2^2 + 2a_1a_2\cos(\Delta \phi)} \]
Substituting the values:
\[ A = \sqrt{3^2 + 3^2 + 2 \cdot 3 \cdot 3 \cdot \cos\left(\frac{\pi}{2}\right)} \]

### Step 5: Calculate the cosine term
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \), the equation simplifies to:
\[ A = \sqrt{3^2 + 3^2 + 0} = \sqrt{9 + 9} = \sqrt{18} \]

### Step 6: Simplify the resultant amplitude
\[ A = \sqrt{18} = 3\sqrt{2} \]

### Final Result
Thus, the amplitude of the resultant wave is:
\[ \boxed{3\sqrt{2}} \]