Question

Two coherent transverse waves of amplitude 2A and A of same frequency propagated respectively along +ve and -ve x-axis are superimposed. The amplitude of the wave at a distance of 5 cm from the origin, $(givenK=\frac{\pi }{4}c{m}^{-1})$ is:

Answer 1

$y_1=A\sin (kx-\omega t)=A\sin (\frac{5\pi }{4}-\omega t)$

$=A(\frac{-1}{\sqrt{2}}\cos \omega t+\frac{1}{\sqrt{2}}\sin \omega t)$

$y_2=2A\sin (kx+\omega t)=2A\sin (\frac{5\pi }{4}+\omega t)$

$=2A(\frac{-1}{\sqrt{2}}\cos \omega t-\frac{1}{\sqrt{2}}\sin \omega t)$

$y=y_1+y_2=-\frac{3A}{\sqrt{2}}\cos \omega t-\frac{A}{\sqrt{2}}\sin \omega t$

Comparing with $y=A\sin \theta +B\cos \theta$

$y_{max}=\sqrt{A^2+B^2}$

Hence here too-

$y_{max}=\sqrt{{\left(\frac{3A}{\sqrt{2}}\right)}^2+{\left(\frac{A}{\sqrt{2}}\right)}^2}$

$⟹y_{max}=\sqrt{5}A$