Equations of motion for two waves traveling i | vedclass

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(C) The given wave equations are $y_1 = A_1 \sin(\omega t - kx)$ and $y_2 = A_2 \sin(\omega t - kx - 	heta)$,where $A_1 = 2a$ and $A_2 = 2a$.Using th

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$4a \cos(	heta / 2)$

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(C) The given wave equations are $y_1 = A_1 \sin(\omega t - kx)$ and $y_2 = A_2 \sin(\omega t - kx - 	heta)$,where $A_1 = 2a$ and $A_2 = 2a$.Using the principle of superposition,the resultant amplitude $A_R$ for two waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by $A_R = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos \phi}$.Here,$A_1 = A_2 = 2a$ and $\phi = 	heta$.$A_R = \sqrt{(2a)^2 + (2a)^2 + 2(2a)(2a) \cos 	heta}$$A_R = \sqrt{4a^2 + 4a^2 + 8a^2 \cos 	heta} = \sqrt{8a^2(1 + \cos 	heta)}$Using the identity $1 + \cos 	heta = 2 \cos^2(	heta / 2)$:$A_R = \sqrt{8a^2 \cdot 2 \cos^2(	heta / 2)} = \sqrt{16a^2 \cos^2(	heta / 2)}$$A_R = 4a \cos(	heta / 2)$.

Equations of motion for two waves traveling in the same direction are given by $y_1 = 2a \sin(\omega t - kx)$ and $y_2 = 2a \sin(\omega t - kx - \theta)$. The resultant amplitude of the medium particle will be:

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