Three harmonic waves having equal frequency v | vedclass

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(A) Let the amplitude of each wave be $A$. Since intensity $I \propto A^2$,we have $I_0 = kA^2$,where $k$ is a constant.The resultant wave equation is

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$5.8 I_{0}$

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(A) Let the amplitude of each wave be $A$. Since intensity $I \propto A^2$,we have $I_0 = kA^2$,where $k$ is a constant.The resultant wave equation is given by the sum of the three waves:$y = A \sin(\omega t) + A \sin(\omega t - \frac{\pi}{4}) + A \sin(\omega t + \frac{\pi}{4})$Using the trigonometric identity $\sin(x - y) + \sin(x + y) = 2 \sin x \cos y$:$y = A \sin(\omega t) + A [2 \sin(\omega t) \cos(\frac{\pi}{4})]$Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$:$y = A \sin(\omega t) + A [2 \sin(\omega t) \cdot \frac{1}{\sqrt{2}}]$$y = A \sin(\omega t) + \sqrt{2} A \sin(\omega t)$$y = (1 + \sqrt{2}) A \sin(\omega t)$The resultant amplitude $A_R = (1 + \sqrt{2}) A$.The resultant intensity $I = k A_R^2 = k [(1 + \sqrt{2}) A]^2 = k A^2 (1 + \sqrt{2})^2$.Since $I_0 = k A^2$:$I = I_0 (1 + 2 + 2\sqrt{2}) = I_0 (3 + 2 	imes 1.414) = I_0 (3 + 2.828) = 5.828 I_0$.Thus,the resultant intensity is approximately $5.8 I_0$.

Three harmonic waves having equal frequency $v$ and same intensity $I_{0}$,have phase angles $0, \frac{\pi}{4}$ and $-\frac{\pi}{4}$ respectively. When they are superimposed,the intensity of the resultant wave is close to

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