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Standing waves problem

  1. Apr 24, 2014 #1
    I'm unclear about the entire idea of why if a string fixed at 2 ends is plucked at a point, say L/3 from the left, the node at the pluck point and any node that is a multiple integer of 3 would be unexcited.

    Trying to see it mathematically, and I've only been able to arrive at is this.

    mL=nλ

    At (1)(L/3), we have L/3 = 3λ

    At (2)(L/3), we have 2L/3 = 6λ

    and so on.

    So, if the string is plucked at L/3, the only wave length that can be arised on the string are n-multiplers of λ.

    3λ, 6λ, 9λ....etc.
     
    Last edited: Apr 24, 2014
  2. jcsd
  3. Apr 24, 2014 #2

    Simon Bridge

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    The standing wave is going to be a skewed triangle in form right?

    Why not sketch out the triangle, then sketch out the next dozen harmonics under it.
    Which harmonics will help sum up to the triangle shape?

    Clearly anything with an antinode close to the peak of the triangle right?
    But anything with a node where the triangle apex has to go will make the sum sort-of dip there and you don't want that.

    Put another way:
    If you added up the multiples-of-3 harmonics, what shape do you get?
     
  4. Apr 24, 2014 #3

    sophiecentaur

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    If you simply pluck a string anywhere but the centre, I think you will not eliminate any of the overtones. You will, of course, alter the spectrum of the sound. You need to partially stop the string and force a node (as when you sound a 'harmonic' on a guitar) - hence eliminate a number of overtones plus the fundamental. Another way to doctor the sound would be to pluck in two places simultaneously. That could eliminate some of the overtones.
     
  5. Apr 24, 2014 #4
    Hi Simon,

    I've seen something like that on a website but unable to wrap my head around still.

    Do you think a mathematical reasoning could be shown?
     
  6. Apr 24, 2014 #5
    If you pluck on a overtone's node, that overtone is not excited.
     
  7. Apr 24, 2014 #6

    sophiecentaur

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    The string can only vibrate in its normal modes and these are found by the discrete Fourier Transform of the shape of the string when released. It is only when the triangle has its apex half way along that there are no even terms, afair.
    This is quite a good link about the topic. I found a couple of other links by my Java wouldn't allow them to run.
     
  8. Apr 24, 2014 #7

    sophiecentaur

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    Oh yes. You are right about that.
     
  9. Apr 24, 2014 #8

    Simon Bridge

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    sophiecentaur's link should work for that.
    But you should be able to do it yourself - any shape wave can be constructed by a sum of sine waves. It's called a Fourier transform.

    You know how to do addition already.
    But you should be able to visualize it -

    Consider the opposite: say youwant to excite the second overtone.

    The third overtone has a node at L/3. That means that this point on the string is not allowed to move all that much (if it moves a lot then, by definition, it is not a node).

    The point that you pluck the string is the point that moves the most.

    Therefore - to excite the 2nd overtone, you cannot pluck the string at x=L/3.

    Therefore, plucking the string at x=L/3 cannot excite the 2nd overtone - or, by extension, any overtone with a node at x=L/3.

    Like I said - draw it out, and do some adding up.
     
  10. Apr 25, 2014 #9

    sophiecentaur

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    Remember, it would be a Discrete Fourier Transform because it is not from -∞ to ∞. This is why there are discrete wavelengths associated with it and not a continuum.
     
  11. Apr 26, 2014 #10
    For standing waves there are 2 equations: mλ/2 and (m+1/2)λ. Are they different?(they shouldn't be!) and when is one used over the next? Can someone show a derivation that explicitly shows the relationship between the first and second?
     
  12. Apr 26, 2014 #11

    sophiecentaur

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    The boundary conditions dictate whether you get a node (closed end on a pipe) or an anti node ( open end). That governs how many half waves there will be.
     
  13. Apr 26, 2014 #12

    Simon Bridge

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    Beware:- what you have written are not equations. Equations have an equals-sign in them.

    mλ/2 = (m+1/2)λ when m=-1.

    But in context I think you mean the conditions for standing waves of the mth overtone.

    (1) L=mλ/2
    (2) L=(m+1/2)λ
    ... m=0,1,2,3...

    If so then these are different conditions, and apply to different situations.

    Why not?

    Determined by the boundary conditions.
    The second one is for one end free to move
    The first is for both end free or both ends fixed.
    There is a third situation, where the "wire" forms a loop.

    You are assuming here that the two equations say the same thing, but they do not. So there is no derivation that will do that.

    You can easily find derivations and examples online in any page devoted to standing waves.
     
    Last edited: Apr 26, 2014
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