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Superposition of harmonic oscillations

  1. Mar 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the amplitude and phase shift of the following two superposed harmonic oscillations.


    2. Relevant equations
    x1(t)=3sin(2∏t+∏/4)
    x2(t)=3cos(2∏t)


    3. The attempt at a solution
    Ok normally i would be able to do this, however one oscillation is cos and the other sin, so i can't use the trig identity for sin+sin or cos+cos. Is it possible to turn 3sin(2∏t+∏/4) into 3cos(2∏t+∏/4+∏/2) or is that invalid?
     
  2. jcsd
  3. Mar 22, 2012 #2
    I think you mean -π/2 there, but otherwise yep, that's fine. (Or there are similar trig identities you could use, but of course it will add up to the same thing.)
     
  4. Mar 22, 2012 #3
    Yeh -∏/2 is what i meant, one more question when calculating the amplitude i find that somehow the following equation is derived, A=√(A12+A22+2A1A2cos(θ21)). I don't understand how this is found?
     
  5. Mar 23, 2012 #4
    Bump: How do I find the amplitude, i get a phase of pi/8 and an amplitude of 5.5 but i think my amplitude is wrong, can someone check please?
     
  6. Mar 23, 2012 #5

    gneill

    User Avatar

    Staff: Mentor

    The formula above is an application of the cosine law. You can see how it applies if you consider that since both have the same frequency ##(2\pi)##, the two functions can be represented in phasor form and added as phasors.

    Convert the first sin function to cosine as you have suggested previously. That makes its phase angle ##-\pi/4##. Its phasor is then a vector of length 3 with angle ##-\pi/4## from the x-axis. The second function has no phase, so it's a vector of length 3 on the x-axis. Add as you would any two vectors. You'll find that one way to find the magnitude is to apply the cosine law -- draw it, you'll see. You could also add them via the component method. Either works just fine.
     
  7. Mar 23, 2012 #6
    Ah, that helps alot thanks very much
     
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