When two progressive waves $y_1=4 \sin (2 x-6 t) \quad$ and $y_2=3 \sin \left(2 x-6 t-\frac{\pi}{2}\right)…
When two progressive waves $y_1=4 \sin (2 x-6 t) \quad$ and $y_2=3 \sin \left(2 x-6 t-\frac{\pi}{2}\right) \quad$ are superimposed, the amplitude of the resultant wave is
Solution
Phase difference between the two waves is $90^{\circ}$. Amplitudes are added by vector method.
$\therefore$ Answer is 5 .