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Superposition of 2 waves problem

  1. Sep 15, 2004 #1
    I am having trouble with this question. Any help would be appreciated.

    Q. Determine the amplitude and phase of the resultant motion when two sinusoidal motions having the same frequency and traveling in the same direction are combined, if their amplitudes are 3.0cm and 4.0cm, and they differ in phase by pi/2 radians.

    We have been covering wave interference, and how they either constructively or destructively interfere.

    What I believe I need to find the new amplitude is:

    A'= sqrt(A1^2 + A2^2 + 2*A1*A2*cos(phi2 - phi1))

    Letting A1=4, A2=3, phi1=0 and phi2= pi/2 radians

    Then I calculate A' as 5cm. I think this is correct. In fact, I graphed the sum of the two equations, Y3 = Y1+Y2 and the max amplitude verifies as 5cm.

    Now I am having trouble with the phase.

    I am trying to use:

    tan(phi3) = (A1*sin(phi1) + A2*sin(phi2)) / (A1*cos(phi1) + A2*cos(phi2))

    Using the same assignments above for the A1, A2 etc...

    tan(phi3) = (4sin(0) + 3sin(pi/2)) / (4cos(0) + 3cos(pi/2))
    tan(phi3) = (0 + 3) / (4 + 0)
    phi3 = arctan(3/4)
    phi3 = .643501pi radians ?

    Is this correct? Am I writing this wrong? I know this is basic, but I am blanking out, and having trouble here.

    Also, the text we are using does not address the superposition of two waves with different amplitudes and a phase shift. The prof's notes are impossible for me to follow. I even broke out my precalc text, and only confused myself further.

    Please help.
  2. jcsd
  3. Sep 15, 2004 #2


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    If the two waves differ in phase by pi/2 then why don't you just take one to be a sine and the other a cosine? That will simplify your calculations quite a bit.

    E.g. if you have A cos x + B sin x then the amplitude is A' sqrt(A^2+B^2) and the new phase is simply phi = -atan(B/A). (The combined wave is A' cos(x + phi))
  4. Sep 16, 2004 #3
    I see what you mean on the new amplitude. The prof gave us that near the end of the lecture, and he also showed us that:

    If A'cos(phi)=A

    Then tan(phi')=B/A
    and phi' can then be calculated.

    But what is the phi used in the sin and cos functions of these arguements?
    That is why I didn't try this approach. I basically don't know how to apply it. I have been bouncing it around all morning, but I am hitting dead ends.
    Can you provide some more guidance?
  5. Sep 16, 2004 #4


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    The basic idea is this:

    Starting with A cos x + B sin x divide by S = sqrt(A^2 + B^2) (of course you have to compensate by multiplying by S which becomes the amplitude!). Now simply define A/S = cos phi and B/S = sin phi. (You can think of A and B as the sides of a right triangle so S is the hypotenuse and the phi is one of the angles of the right triangle).

    Once you have that then A cos x + B sin x = cos phi * cos x + sin phi * sin x and you can now use the obvious trig identity.
  6. Sep 20, 2004 #5
    I am sorry Tide, but I looked at this over the weekend, and I am still not getting this method of solving.

    I am going to submit this problem with the long method I showed above, and my final answers are A'=5cm, and phi'=.644 radians. I hope this is correct.

    However, I would like to fully understand your method, so if you can (or someone else in case Tide is tired of dealing with this), please explain in further detail.

    Specifically, I am not getting your first step:

  7. Sep 20, 2004 #6


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    [tex]A \cos x + B \sin x = \sqrt {A^2 + B^2} \frac {A \cos x + B \sin x}{\sqrt {A^2 + B^2}}[/tex]

    Now define

    [tex]\frac {A}{\sqrt {A^2 + B^2}} = \cos \phi[/tex]
    [tex]\frac {B}{\sqrt {A^2 + B^2}} = \sin \phi[/tex]

    [tex]A \cos x + B \sin x = \sqrt {A^2 + B^2} \left( \cos x \cos \phi + \sin x \sin \phi \right)[/tex]

    Finally, use the trig identity on the last part.
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