Question #23366

a wave of amplitude A and intensity I is coincident with a second wave of amplitude 3A. both waves have the same frequency. calculate in tems of A and c the resultant amplitude and intensity when phase difference is zero and when it is [pie-rad]!

Expert's answer

Question 23366

x1(t)=Acos(ωt)=Acos(2πCλt),I=A2.x _ {1} (t) = A \cos (\omega t) = A \cos \left(2 \pi \frac {C}{\lambda} t\right), \quad I = A ^ {2}.x2(t)=3Acos(ωt+δ)=3Acos(2πCλt+δ),I=9A2.x _ {2} (t) = 3 A \cos (\omega t + \delta) = 3 A \cos \left(2 \pi \frac {C}{\lambda} t + \delta\right), \quad I = 9 A ^ {2}.x(t)=x1(t)+x2(t)=Acos(2πCλt)+3A[cos2πCλtcosδsin2πCλtsinδ]x (t) = x _ {1} (t) + x _ {2} (t) = A \cos \left(2 \pi \frac {C}{\lambda} t\right) + 3 A \left[ \cos 2 \pi \frac {C}{\lambda} t \cdot \cos \delta - \sin 2 \pi \frac {C}{\lambda} t \cdot \sin \delta \right]


For δ=0:x(t)=Acos(2πCλt)+3A[cos2πCλt]=4Acos2πCλt,I=16A2.\delta = 0: x(t) = A\cos \left(2\pi \frac{C}{\lambda} t\right) + 3A[\cos 2\pi \frac{C}{\lambda} t] = 4A\cos 2\pi \frac{C}{\lambda} t, I = 16A^2.

For δ=π:x(t)=Acos(2πCλt)3A[cos2πCλt]=2Acos2πCλt,I=4A2.\delta = \pi : x(t) = A\cos \left(2\pi \frac{C}{\lambda} t\right) - 3A[\cos 2\pi \frac{C}{\lambda} t] = -2A\cos 2\pi \frac{C}{\lambda} t, I = 4A^2.

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