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 given waves
The two progressive waves are:
- \( y_1 = 4 \sin(2x - 6t) \)
- \( y_2 = 3 \sin(2x - 6t - \frac{\pi}{2}) \)

### Step 2: Determine the phase difference
The second wave can be rewritten using the sine function's phase shift:
- \( y_2 = 3 \sin(2x - 6t - \frac{\pi}{2}) = 3 \cos(2x - 6t) \)

The phase difference between the two waves is:
- \( \Delta \phi = \frac{\pi}{2} \) (since \( \sin \) and \( \cos \) are 90 degrees out of phase).

### Step 3: Use the formula for resultant amplitude
When two waves with amplitudes \( A_1 \) and \( A_2 \) interfere with a phase difference \( \Delta \phi \), the amplitude \( A \) of the resultant wave can be calculated using the formula:
\[ A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\Delta \phi)} \]

### Step 4: Substitute the values
Here, \( A_1 = 4 \), \( A_2 = 3 \), and \( \Delta \phi = \frac{\pi}{2} \):
- \( \cos\left(\frac{\pi}{2}\right) = 0 \)

Thus, the formula simplifies to:
\[ A = \sqrt{4^2 + 3^2 + 2 \cdot 4 \cdot 3 \cdot 0} \]
\[ A = \sqrt{16 + 9} \]
\[ A = \sqrt{25} \]
\[ A = 5 \]

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

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