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Homework Help: Relating Arc Length and Standing Wave Patterns

  1. Dec 11, 2013 #1
    1. The problem statement, all variables and given/known data

    I am currently reviewing the physics of 'standing waves on a string'. I know that for the nth harmonic, the length of the 'string' is [itex]\frac{n\lambda}{2}[/itex]. Instead of just memorizing these, I have been trying to apply my knowledge of Calculus to figure out why these numbers are what they are. However, I am having a bit of trouble. Here's what I have tried to do thus far:

    3. The attempt at a solution

    If you take the graph of Sin(x), the arc length of the curve between 0 and a can be calculated with the following integral:

    [itex]\int \sqrt{1+cos(x)^{2}}dx[/itex] (with limits 0 and a). So one wavelength would be equal to a distance of 2[itex]\pi[/itex]. The integral between 0 and [itex]\pi[/itex] [itex]\approx 3.82 [/itex] while the integral between 0 and 2[itex]\pi[/itex] [itex]\approx 7.64 [/itex]

    So, that means if we had a wave that was a sine function defined between 0 and [itex]\pi[/itex], the wavelength would be 1/2 that of a sin function between 0 and 2[itex]\pi[/itex] and it would have a length of 1/2 that of the function defined between 0 and 2[itex]\pi[/itex].

    What I am confused about is that I keep just getting:

    First Harmonic [itex]\frac{\lambda}{2}[/itex]=[itex]\frac{L}{2}[/itex] ([itex]\lambda[/itex]=L)

    Apart from it being the wrong answer, this would mean the wavelength, [itex]\pi[/itex], is equal to L which by the arc length formula is 3.82. Furthermore, would the wavelength have to be the same as the arc length (i.e. length of the string?)

    Also if I said I wanted to find the length of the sine wave between 0 and [itex]\pi[/itex] wouldn't I be able to say that the wavelength= [itex]\pi[/itex] and by definition L=wavelength/2 for first harmonic. Then I would get [itex]\pi[/itex]/2 = 3.82. 1.57≠3.82 :grumpy:

    Maybe by Length of the string they don't mean arc length?

    Am I just completely missing something here, or am I using Calculus in the wrong place.


    (EDIT) I think I figure it out. I was making it WAY more complicated then it has to be. Since there are two nodes on each end, it is like fixing a string in two positions and pulling it up, so the length of the string can't change then.
    Last edited: Dec 11, 2013
  2. jcsd
  3. Dec 12, 2013 #2

    Simon Bridge

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    Well done :)

    Basically the model is for those situations where the overall length of the string can change while the tension remains constant and the separation of the endpoints is a constant ... where there are not a lot of losses. If there are no losses, the standing wave does not even have to be a sine wave. (i.e. pluck it in the middle and you have a triangular standing wave.)

    The equation you start with for calculus is the wave equation in one dimension:
    ... the solution is generally not trivial (from scratch) but the upshot is that the solutions can be expanded in a trigonometric series, and you are back to counting nodes and antinodes between 0 and L.

    Aren't you glad you asked ?!
  4. Dec 12, 2013 #3


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    Correct. The wave length λ is the straight line distance between two nodes (or antinodes). The length of string refers to the length with no wave on it.

    Edit: oops my post crossed with Simons.
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