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Standing waves energy problem

  1. May 24, 2005 #1
    The expression T(tension)/(mu*f^2) shows that no matter what the tension, and frequency are, as long as there are the same amount of nodes in the string, the ratio is always the same.

    The reason that the ratio in the previous part always comes out the same, regardless of which of the many possible configurations of the string that give four antinodes you chose, is that the ratio actually equals a much simpler quantity that will always be the same for configurations of the string that yield four antinodes. Which of the following gives that quantity? Here E is the energy of the wave, and A is the amplitude.

    a.lambda b.lambda^2 c.E d.E^2 e.A f.A^2

    I'm not sure what the answer is. Would someone please help.

  2. jcsd
  3. May 24, 2005 #2
    Just by looking at the equation you can tell that it is going to have dimensions of length squared. So the answer is either lambda^2 or A^2. The wave speed in a rope is given by [itex]c = \sqrt{\frac{T}{\mu}}[/itex] which is also equal to [itex]c=\nu\lambda[/itex]. Equating these expressions, you may then write down what [itex]\frac{T}{\mu\nu^2}[/itex] is equal to.
  4. May 24, 2005 #3
    Oh ok. Thanks for that.
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