Two waves Y_1 = A_1 sin(omega t - beta_1) and | vedclass

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(A) The two waves are given by $Y_1 = A_1 \sin(\omega t - \beta_1)$ and $Y_2 = A_2 \sin(\omega t - \beta_2)$.The phase difference between the two wave

Vedclass · Accepted Answer

$\sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\beta_1 - \beta_2)}$

Vedclass · Answer

(A) The two waves are given by $Y_1 = A_1 \sin(\omega t - \beta_1)$ and $Y_2 = A_2 \sin(\omega t - \beta_2)$.The phase difference between the two waves is $\phi = (\omega t - \beta_1) - (\omega t - \beta_2) = \beta_2 - \beta_1$.The resultant amplitude $A$ of two superimposing waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by the formula:$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$.Substituting $\phi = \beta_2 - \beta_1$ into the formula,and noting that $\cos(\beta_2 - \beta_1) = \cos(\beta_1 - \beta_2)$:$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\beta_1 - \beta_2)}$.

Two waves $Y_1 = A_1 \sin(\omega t - \beta_1)$ and $Y_2 = A_2 \sin(\omega t - \beta_2)$ superimpose to form a resultant wave whose amplitude is

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