Two waves y_1 = A_1 sin(omega t - beta_1) and | vedclass

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(A) The phase difference between the two waves is given by $\phi = (\omega t - \beta_2) - (\omega t - \beta_1) = \beta_1 - \beta_2$.According to the p

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

Vedclass · Answer

(A) The phase difference between the two waves is given by $\phi = (\omega t - \beta_2) - (\omega t - \beta_1) = \beta_1 - \beta_2$.According to the principle of superposition,the resultant amplitude $A$ of two 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 the value of $\phi$,we get $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\beta_1 - \beta_2)}$.

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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