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(A) The given wave equations are ${y}_{1} = {A}_{1} \sin {k}({x} - {vt})$ and ${y}_{2} = {A}_{2} \sin {k}({x} - {vt} + {x}_{0})$.The phase difference

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$7$

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(A) The given wave equations are ${y}_{1} = {A}_{1} \sin {k}({x} - {vt})$ and ${y}_{2} = {A}_{2} \sin {k}({x} - {vt} + {x}_{0})$.The phase difference $\Delta \phi$ between the two waves is given by $\Delta \phi = {k} \cdot {x}_{0}$.Given ${k} = 6.28 \, {cm}^{-1}$ and ${x}_{0} = 3.5 \, {cm}$.Since $6.28 \approx 2\pi$,we have $\Delta \phi = 6.28 	imes 3.5 = 2\pi 	imes 3.5 = 7\pi$.The resultant amplitude ${A}_{R}$ is given by ${A}_{R} = \sqrt{{A}_{1}^{2} + {A}_{2}^{2} + 2{A}_{1}{A}_{2} \cos(\Delta \phi)}$.Substituting the values: ${A}_{R} = \sqrt{12^{2} + 5^{2} + 2(12)(5) \cos(7\pi)}$.Since $\cos(7\pi) = -1$,we get ${A}_{R} = \sqrt{144 + 25 - 120} = \sqrt{49} = 7 \, {mm}$.

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