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Math question from the inumerate

  1. Jul 18, 2003 #1
    Can someone tell me how to do this.

    How can I measure the length of the actual curve of a sine wave. The actual curve not the wavelength.

    Raavin :)
     
  2. jcsd
  3. Jul 18, 2003 #2
    well, you just have to use the arc length formula.

    for a function f(x),

    arc length S =[inte] [squ](1+ [f'(x)]2 )dx

    and then of course you just need to integrate and evalute it for your boundries.
     
  4. Jul 18, 2003 #3
    It would go something like this
    y=sin(x)

    dl^2=dx^2+dy^2 (Pythagorean theorem applied at the infinitesimal scale where dl is the length of the infinitesimal segment of the curve)

    dl=√(dx^2+dy^2)
    dl=√(1+dy^2/dx^2)dx
    l=∫0…2π √(1+(dy/dx)^2)dx
    l=∫0…2π √(1+cos(x)^2)dx
    l=6.28 (approximatly)

    Here is a link that may help
    http://mathworld.wolfram.com/ArcLength.html
     
  5. Jul 18, 2003 #4
    no, 6.28 is 2 pi, the circumference of a circle of radius 1

    the sin curve, from 0 to pi has a length of approximently 3.8201977890278
     
  6. Jul 20, 2003 #5
    You are correct on the interval from 0 to π the length is as you stated but my answer is for [0,2π]

    I assumed he wanted the length of a single wave which is produce over the interval [0,2π]
     
  7. Jul 20, 2003 #6
    suicidal, i think you should check your numbers, ORW is correct.
     
  8. Jul 20, 2003 #7
    I'm trying to figure out what the relationship of the amplitude of waves of different frequencies given a fixed 'arc length?' and the total 'spread' being the same. So a standing wave with no elasticity. I'll try the formulas to see if I can work it out. If someone knows the relationship already then feel free to share also.

    Raavin :)
     
  9. Jul 20, 2003 #8
    Could you explain this to me like I'm an idiot???

    Explaining what I said above, I want to be able to create a sine wave with one 'period??', calculate the arc length, then create another wave with two periods with the same boundry and the same arc length. So the amplitude will be less on the second one, then do the same with 3 periods, 4 periods etc. Then I want to see if the relationship of the amplitudes corresponds to any other physics relationships.

    Get what I mean???

    Raavin :)
     
  10. Jul 20, 2003 #9

    Hurkyl

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    You don't need to measure the length of a sine wave to do that. :smile: You just need similarity.

    If you scale a figure by a ratio r, all lengths in the figure are multiplied by r.

    So suppose our initial sine wave has one period in [0, 2π], amplitude 1, and an unknown arclength L.

    If we scale the figure by one half, it will have one period in [0, π], amplitude 1/2, and an unknown arclength L/2.

    So, if we put two copies of the diminished figure together, we get two periods in [0, 2π], amplitude 1/2, and unknown arclength L.

    Similarly if we scaled by 1/3; you'd fit 3 periods into [0, 2π] with an amplitude of 1/3 and arclength L.



    (Note: the theorem for scaling only works if you scale the same amount in all directions; if you, say, stretch in the x-direction but leave the y-direction unchanged, there isn't an easy formula for lengths)
     
  11. Jul 21, 2003 #10
    i found my mistake. i evaluated the integral incorrectly.

    i used a calculator.
    is there a way to evaluate ∫√(1+cos(x)^2)dx by hand
     
  12. Jul 21, 2003 #11
    this is an elliptic integral, so, no, not really.
     
  13. Jul 21, 2003 #12
    Ok, the scale thing makes perfect sense. I don't think I'm going to get the result I was looking for but another question. Unfortunately as I said, my maths is poor. Just going on intuition however, if you scale by half,including the length, the area under the curve reduces to a quarter right???

    But if you keep the same total length, then the area under the curve reduces by the same proportions as the amplitude and the wavelength? eg. halve the wavelength and amplitude and you halve the area under the curve because you are dividing by four and multiplying by 2.

    Is that right. If it is, what happens when you reduce the length by less than a full wavelength or reduce the wavelength and amplitude by a non integer amount. Does this stuff the scaling???

    Raavin
     
  14. Jul 21, 2003 #13

    Hurkyl

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    If you scale a figure by r, all linear functions of size (such as length) are scaled by r. All quadratic functions of size (such as area) are scaped by r2. All cubic functions (like volume) by r3, et cetera.

    So yes, if you scale by 1/2, all areas are multipled by 1/4.

    And since you take two copies of the scaled figure, the resulting area would be half as much as the original.



    You can scale the figure by any real number, but as you noted you might not be able to fit an integral number of scaled periods into your original period... and there's no reason to think that the arclength or area under the sine wave should be proportional to the (fractional) number of copies because the replications aren't complete.
     
  15. Jul 21, 2003 #14
    Ok, so we scale so that 2.5 or even 2.25 waves fit, it would seem logical that we get the same ratio of area scaling because we have the extra full half wave or part wave with a symmetrical division. The problem arises when we don't have a full half because for example the area of the 1st 1/4 of a node is not 1/4 of the area of the full node (is node the right term??). So during the scaling period in between the discreet 1/4 steps there is incongruence in area and length scaling.

    The reason I'm interested in this is in trying to find a simple connection between this effect and discreetness in quantum physics.

    This is probably pushing it, but if one was to plot the variation in the area to length ratio in this case, would it fit a standard function????

    Raavin [?]
     
    Last edited: Jul 21, 2003
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