When two progressive waves `y_(1) = 4 sin (2x - 6t)` and `y_(2) = 3 sin (2x - 6t - (pi)/(2))` are superimposed, the amplitude of the resultant wave is

Explanation

To find the amplitude of the resultant wave when two progressive waves are superimposed, we can follow these steps:

### Step 1: Identify the Amplitudes and Phase Difference
Given the two waves:
- \( y_1 = 4 \sin(2x - 6t) \)
- \( y_2 = 3 \sin(2x - 6t - \frac{\pi}{2}) \)

From these equations, we can identify:
- Amplitude of the first wave, \( a_1 = 4 \)
- Amplitude of the second wave, \( a_2 = 3 \)

Next, we need to find the phase difference \( \phi \) between the two waves. The phase of the second wave can be rewritten as:
- \( y_2 = 3 \sin(2x - 6t - \frac{\pi}{2}) = 3 \cos(2x - 6t) \)

This indicates that the phase difference \( \phi \) is \( -\frac{\pi}{2} \).

### Step 2: Use the Formula for Resultant Amplitude
The formula for the resultant amplitude \( A \) when two waves with amplitudes \( a_1 \) and \( a_2 \) and a phase difference \( \phi \) are superimposed is given by:
\[ A = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos(\phi)} \]

### Step 3: Calculate the Cosine of the Phase Difference
Since the phase difference \( \phi = -\frac{\pi}{2} \):
\[ \cos\left(-\frac{\pi}{2}\right) = 0 \]

### Step 4: Substitute Values into the Amplitude Formula
Now, substituting the values into the formula:
\[ A = \sqrt{4^2 + 3^2 + 2 \cdot 4 \cdot 3 \cdot 0} \]
\[ A = \sqrt{16 + 9 + 0} \]
\[ A = \sqrt{25} \]
\[ A = 5 \]

### Conclusion
The amplitude of the resultant wave is \( 5 \).

---