Wave Superposition: Solving for Resultant Amplitude

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Two waves with amplitudes of 1.26 and 2.33 interfere with a phase difference of 62 degrees, leading to a resultant amplitude calculation. The initial attempt at calculating the amplitude using a sine function was incorrect, as it did not account for the phase difference correctly. The correct method involves representing the waves as phasors and adding them vectorially, which clarifies the resultant amplitude. The discussion highlights the importance of using phasor addition to solve interference problems accurately. The final understanding of the concept was achieved through this phasor explanation, aiding in exam preparation.
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Homework Statement


Two waves of the same frequency have amplitudes of 1.26 and 2.33. They interfere at a point where their phase difference is 62.0(deg). The resultant amplitude is:
A. 3.13
B. 3.59
C. 0.77
D. 1.21
E. 2.15


Homework Equations


y(x,t) = y_m sin(kx-wt)


The Attempt at a Solution


Well I know right away the answer is not B because there is a phase difference. So what I did was (1.26+2.33)*sin(62) and I get 3.17 which is not in the list. I am supposed to just assume its 3.13 (A) or am I doing something wrong. Thanks help is appreciated.
 
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Represent the waves as phasors.
Add the phasors head-to-tail just like vectors.
The length of the result is the amplitude you are after.

eg. if they were the same amplitude, but the relative phase was exactly 60 degrees, then the resulting vector would be 3A/2 by √3.A/2 (pythagoras) gives √3.A as the amplitude. (The net phase is just arctan(1/√3).)
 
YES! thank you I got it now... thanks for the explanation the phasors helped me figure it out, forgot that little bit from lecture. I am reviewing for a final your a life saver.
 
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