Amplitude of Resultant Wave from Superposition

To find the amplitude of the resultant wave when two progressive waves superimpose, we can use the principle of superposition.

The given wave equations are:
$y_1 = 4 \sin(2x - 6t)$
$y_2 = 3 \sin\left(2x - 6t - \frac{\pi}{2}\right)$

From these equations, the amplitudes of the individual waves are:
$A_1 = 4$
$A_2 = 3$

The phase difference $\Delta \phi$ between the two waves is:
$\Delta \phi = (2x - 6t) - \left(2x - 6t - \frac{\pi}{2}\right) = \frac{\pi}{2}$

The amplitude $A$ of the resultant wave formed by the superposition of two waves with amplitudes $A_1$ and $A_2$ and phase difference $\Delta \phi$ is given by the formula:
$A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\Delta \phi)}$

Substituting the values into the formula:
$A = \sqrt{4^2 + 3^2 + 2(4)(3) \cos\left(\frac{\pi}{2}\right)}$

Since $\cos\left(\frac{\pi}{2}\right) = 0$:
$A = \sqrt{16 + 9 + 0} = \sqrt{25} = 5$

Thus, the amplitude of the resultant wave is $5$.