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Homework Help: Amplitude of Two Waves in Interference

  1. Oct 28, 2012 #1
    In my textbook, when two waves of the form:
    [tex] y_1 = A\cos \left( 2\pi f_1t \right)[/tex]
    [tex] y_2 = A\cos \left( 2\pi f_2t \right)[/tex]
    combine, the following trig identity is used:
    [tex] \cos a + \cos b = 2\cos \left( \dfrac{a-b}{2} \right) \cos \left( \dfrac{a+b}{2} \right) [/tex]
    which yields an expression for y:
    [tex]y=\left[ 2A\cos 2\pi \left( \dfrac{f_1-f_2}{2} \right) t \right] \cos 2\pi \left( \dfrac{f_1+f_2}{2} \right) t [/tex]
    and thus the Amplitude for the resultant wave is the expression in the square brackets. BUT...why can't the order be switched, yielding:
    [tex]y=\left[ 2A\cos 2\pi \left( \dfrac{f_1+f_2}{2} \right) t \right] \cos 2\pi \left( \dfrac{f_1-f_2}{2} \right) t [/tex]
    Which seems to be a different wave with a different amplitude... What's going on here? Why am I forced to use this expression for the amplitude rather than the other one?
  2. jcsd
  3. Oct 28, 2012 #2


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    Hi Bennigan88! :smile:
    They can be switched, it's the same.

    Since the - is presumably a lot smaller than the +,

    the graph looks like a very zigzaggy line going up and down inside a cos curve corresponding to the + …

    "from a distance", the amplitude looks just like the amplitude of the + :wink:
  4. Oct 28, 2012 #3
    So then what about this question: Two waves are described by y1 = 6 cos 180t and y2 = 6 cos 186t (both in meters). With what angular frequency does the maximum amplitude of the resultant wave vary with time?

    Resultant wave:
    [tex]y_1+y_2=6\left(\cos 180t + \cos 186t \right) [/tex]
    [tex]=12\cos\left(\dfrac{366}{2}t\right)\cos\left( \dfrac{6}{2}t\right)[/tex]
    [tex]=12\cos \left(183t\right)\cos \left(3t\right)[/tex]

    So...what is the question asking me? If the amplitude is 12cos183t then the answer is 183, but if the amplitude is 12cos3t, then the answer is 3. I'm still confused. Also, the answer is supposed to be 3 rad/s.
  5. Oct 28, 2012 #4


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    See attachment, it is your wave. Does it look a wave with its amplitude changing with angular frequency ω=3 rad/s or a wave with amplitude varying with ω=183 rad/s?


    Attached Files:

  6. Oct 29, 2012 #5
    Thank you for the graph but that does not explain why one term defines the change in amplitude and not the either.

    If you looked at the higher angular frequency curve as defining the amplitude, because the other factor has such a lower angular frequency it doesn't "fill up" the the other curve, in fact you can't even detect it, so in general the factor with the lower angular frequency defines the amplitude of the resultant wave?
  7. Oct 29, 2012 #6


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    The low frequency curve is the envelope of the high frequency one. During one period of the lower frequency curve (the red one) there are a lot of maxima and minima of the high-frequency wave. But these maxima and minima change with time. We call "amplitude" the extremal deviation from zero. You can define amplitude for the high frequency curve during any one of its periods, but it is impossible to find a single minimum and maximum during one period of the low-frequency curve. That is why we say that the amplitude of the high frequency curve changes.

    To consider the product of two waves as one wave with varying amplitude is justified only in case when the frequencies are very different in magnitude.

  8. Oct 29, 2012 #7
    Thank you, I think that answers the question pretty well.
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