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Produce Standing Wave Problem

  1. Apr 28, 2015 #1
    1. The problem statement, all variables and given/known data
    The equation of a transverse wave traveling in a string is given by y(x,t) = 10 cos (π/2)(0.0050x - 8.0t + 0.57), in which x and y are expressed in centimeters and t in seconds. Write down the equation of a wave which, when added to the given one, would produce standing waves on the string.

    2. Relevant equations
    y'(x,t) = [2ymcos(0.5β)]sin(kx - ωt+ 0.5β)
    y'(x,t) = [2ym sin(kx)]cos(ωt)

    3. The attempt at a solution
    The problem is...I don't know what to do with that phase constant in the way. Like, I know that if the equation was just:

    y(x,t) = 10 cos (π/2)(0.0050x - 8.0t)

    I would just have to go through some really, really obnoxious trig to figure it out. But I am at a complete loss here. How do I even figure it out with that stupid phase constant around to muck things up?
     
  2. jcsd
  3. Apr 28, 2015 #2

    haruspex

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    Don't worry about the phase constant for the moment. If a wave is reflected without losing amplitude then a standing wave will result, yes? In the wave equation ##A \sin(k x -\omega t +\phi)##, what determines which way the wave is moving? How would you change it into a wave moving at the same speed, with the same wavelength, same amplitude, but in the opposite direction?
     
  4. Apr 28, 2015 #3

    andrevdh

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    A standing wave is formed by the superposition of two identical waves travelling in opposite directions.
    The phase constant just causes a constant offset, or shift, in the wave with respect to a wave with a zero
    phase constant.
     
  5. Apr 28, 2015 #4
    The omega*t determines which way it's moving, right? That's why waves traveling in the negative direction have the equation
    ##A \sin(k x +\omega t +\phi)##
    right?
     
  6. Apr 28, 2015 #5
    So, I attempted to get some help from my friend. She screwed up a bit during the process (basically, she didn't alter the direction in which the wave was moving, so basically added the exact same equation to the original one) but:

    y(x,t) = 10 cos (π/2)(0.0050x - 8.0t - 0.57)
    I think I messed up and wrote (+) here. We're actually subtracting the phase constant; that was a typo

    Same wave, just in the opposite direction:
    y(x,t) = 10 cos (π/2)(0.0050x + 8.0t - ϕ)

    So:
    y' = 10 cos (π/2)(0.0050x - 8.0t - 0.57) + 10 cos (π/2)(0.0050x + 8.0t - ϕ)

    She said this was essentially:
    y' = 10 cos (π/2)(U) + 10 cos (π/2)(V)

    And from there, you could do the trig identity:
    y' = 2*10*cos(π/2)(0.5(U+V))*cos(π/2)(0.5(U-V))
    y' = 20*cos(π/2)(0.5(2*0.0050x - 0.57 - ϕ)*cos(π/2)(0.5(-16.0t - 0.57 + ϕ))

    So...is that it? Is that as far as you can get? Is the phase constant actually solvable?
     
  7. Apr 28, 2015 #6

    andrevdh

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    The original question did not require you to actually go this far.
    You just need to state the 2nd required wave equation.
    You also should not change the sign of the phase constant, that is keep it + phi.
     
  8. Apr 28, 2015 #7
    No. The original equation (I had a typo when I transcribed it from my homework) it was - phi.
    So the second required wave equation is:
    10 cos (π/2)(0.0050x + 8.0t - ϕ).

    How do you find phi?
     
  9. Apr 28, 2015 #8

    haruspex

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    Half right. Moving to the right would mean that increasing t looks the same as increasing x. What does that tell you about the relationship between k and omega?
    There are no boundary conditions specified, it just says a standing wave. The phase of the standing wave doesn't matter.
     
  10. Apr 29, 2015 #9

    andrevdh

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    What are the other questions that you need to address?

    The reasoning behind determining which way the wave is moving works like this:

    In the wave equation we have that its phase is the term (kx ± ωt ± Φ).
    In order to figure out which way the wave is moving we need to "watch" a particular
    part of the wave as the wave progresses, say for instance we look at a particular
    antinode. Both the x and t values will change as the wave moves through the medium,
    but the displacement of the particular antinode , y, that we are observing stays the same.
    In order for this to happen the value of the phase need to stay the same, or be a specific
    constant value, say C. So for this antinode we have that

    C = kx ± ωt ± Φ

    Now if we increase time or t we have to alter x in order to keep the phase constant.
    From this info we can see whether the x value need to increase or decrease in order
    to keep C constant, which tells us which way the wave is moving.
     
    Last edited: Apr 29, 2015
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