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Standing Waves

  1. Sep 27, 2006 #1
    Hi.
    Okay, this has been driving me crazy. When combining two given waves into a standing wave equation, how do you know which sign to put in front of the amplitude? All the examples I've been finding seem to contradict each other. Here are three examples from my textbook:

    1. Incoming Wave:
    y=4sin[2pi(t)-6pi(x)]

    Reflected Wave:
    y=-4sin[2pi(t)+6pi(x)]

    Standing Wave:
    y=-8cos3(pi)t*sin6(pi)x

    ---------------------------

    2. Incoming Wave
    y=-4sin[2pi(t)+6pi(x)]

    Reflected Wave:
    y=4sin[2pi(t)-6pi(x)]

    Standing Wave:
    y=8cos3(pi)t*sin6(pi)x

    ---------------------------

    3. Incoming:
    y=-8sin[2(pi)t-7(pi)x]

    Reflected Wave:
    y=8sin[2(pi)t+7(pi)x]

    Standing:
    y=16cos2(pi)t*sin7(pi)x
    ---------------------------

    Help?!?
    How do I know when it's positive or when it's negative?
     
  2. jcsd
  3. Sep 27, 2006 #2

    quasar987

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    Gold Member

    Well there is at least one common denominator to all there cases. In each of them, the reflected wave's amplitude is of opposite sign to the incoming one. The resulting amplitude in the standing wave is a result of algebra. For exemple, for the first:

    1. Incoming Wave:
    y=4sin[2pi(t)-6pi(x)]

    Reflected Wave:
    y=-4sin[2pi(t)+6pi(x)]

    Standing = Incoming + Reflected = 4sin[2pi(t)-6pi(x)] - 4sin[2pi(t)+6pi(x)] = 4{sin[2pi(t)-6pi(x)]+sin[-2pi(t)-6pi(x)]}

    I used the fact that -sin(x) = sin(-x). Now I'll use the identity sin(A-B)+sin(A+B)=2sinAcosB with A= -6pi(x) and B=-2pi(t):

    Standing = 8sin[-6pi(x)]cos[-2pi(t)] = -8sin[6pi(x)]cos[2pi(t)]

    I used again the identity -sin(x) = sin(-x) as well as cos(x) = cos(-x).
     
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