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Superposition of Waves - Standing Waves

  1. Nov 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider a situation in which a wave is traveling in the negative x-direction encounters a barrier and is reflected. Assume an ideal situation in which none of the energy is lost on reflection nor absorbed by the transmitting medium. This permits us to write both waves with the same amplitude. I will represent these equations as

    [itex]E_{1} = E_{0}sin(ωt + kx)[/itex]
    [itex]E_{2} = E_{0}sin(ωt - kx - θ_{R})[/itex]

    Here [itex]θ_{R}[/itex] is included to account for possible phase shifts upon reflection. The resultant wave of the two waves can be represented as

    [itex]E_{R} = E_{1} + E_{2} = E_{0}[sin(ωt + kx) + sin(ωt - kx - θ_{R})][/itex]

    Next I make the substitution

    [itex]β_{+} = ωt + kx[/itex] and [itex]β_{+} = ωt - kx - θ_{R}[/itex]

    and employ the identity

    [itex]sin(β_{+}) + sin(β_{-}) = 2sin(\frac{1}{2}(β_{+} + β_{-}))cos(\frac{1}{2}(β_{+} + β_{-}))[/itex]

    This yields

    [itex]E_{R} = 2E_{0}cos(kx + \frac{θ_{R}}{2})sin(ωt - \frac{θ_{R}}{2})[/itex]

    Consider the situation in which a standing wave results when [itex]\frac{θ_{R}}{2} = \frac{∏}{2}[/itex] and you get

    [itex]E_{R} = 2E_{0}sin(kx)cos(ωt)[/itex]

    2. Relevant equations

    3. The attempt at a solution

    This is what my book claims. The only problem I have is that it looks like it made the substitution

    [itex]sin(x - \frac{∏}{2}) = -cos(x)[/itex] and [itex]cos(x - \frac{∏}{2}) = sin(x)[/itex]

    The problem is that when I make these substitutions I get


    I'm not exactly sure how it's supposed to be positive. Thanks for any help.
  2. jcsd
  3. Nov 26, 2013 #2
    if cos(x-pi/2) = sin(x), what does cos(x+pi/2) equal?
  4. Nov 26, 2013 #3
    Oh wow. I can't believe I didn't see that thanks.
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