The amplitude of the resultant wave produced | vedclass

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(A) The resultant wave is $y = y_1 + y_2$.$y = A_1 \sin(wt - \beta_1) + A_2 \sin(wt - \beta_2)$.Using the identity $\sin(x - y) = \sin x \cos y - \cos

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$[A_1^2 + A_2^2 + 2A_1A_2 \cos(\beta_1 - \beta_2)]^{1/2}$

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(A) The resultant wave is $y = y_1 + y_2$.$y = A_1 \sin(wt - \beta_1) + A_2 \sin(wt - \beta_2)$.Using the identity $\sin(x - y) = \sin x \cos y - \cos x \sin y$:$y = A_1(\sin wt \cos \beta_1 - \cos wt \sin \beta_1) + A_2(\sin wt \cos \beta_2 - \cos wt \sin \beta_2)$.Grouping the terms:$y = (A_1 \cos \beta_1 + A_2 \cos \beta_2) \sin wt - (A_1 \sin \beta_1 + A_2 \sin \beta_2) \cos wt$.Let $X = A_1 \cos \beta_1 + A_2 \cos \beta_2$ and $Y = A_1 \sin \beta_1 + A_2 \sin \beta_2$.The resultant amplitude $A$ is $\sqrt{X^2 + Y^2}$.$A = \sqrt{(A_1 \cos \beta_1 + A_2 \cos \beta_2)^2 + (A_1 \sin \beta_1 + A_2 \sin \beta_2)^2}$.Expanding the squares:$A = \sqrt{A_1^2(\cos^2 \beta_1 + \sin^2 \beta_1) + A_2^2(\cos^2 \beta_2 + \sin^2 \beta_2) + 2A_1A_2(\cos \beta_1 \cos \beta_2 + \sin \beta_1 \sin \beta_2)}$.Using trigonometric identities $\sin^2 	heta + \cos^2 	heta = 1$ and $\cos(A - B) = \cos A \cos B + \sin A \sin B$:$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\beta_1 - \beta_2)}$.

The amplitude of the resultant wave produced by the superposition of two waves $y_1 = A_1 \sin(wt - \beta_1)$ and $y_2 = A_2 \sin(wt - \beta_2)$ is:

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