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Wave equations clarification

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data

    L = [(2n -1) / 4 ] lamda
    I dont understand what is this

    and also L = n lamda / 2

    In my notes
    it only says
    f = c/ 4L for 1st harmonic
    f = 3c / 4L for 2nd harmonic.. etc

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 16, 2013 #2
    the lamda and L are very confusing to me..
     
  4. Apr 16, 2013 #3

    haruspex

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    L is the length of some physical object capable of standing wave modes of vibration. Lambda is the wavelength (within the object) of one of these modes. As the object vibrates, some points stay still (antinodes) while the points halfway between these (nodes) undergo the maximum amplitude of vibration. A complete wavelength is twice the distance between consecutive nodes.
    How many wavelengths fit in L depends on :
    a) are the ends nodes or antinodes?
    b) how many 'extra' wavelengths are in L
    Suppose the object is a pipe open at one end only. (The end with the reed should be considered open.) The open end is a node, while the closed end is an antinode - the air cannot move much there. The simplest possibility is just the one node and the one antinode, making L one quarter of a complete wavelength. This is the n=1 case of your formula. (Try and sketch this.) This mode is the fundamental, or 'first harmonic'.
    The next possibility has an extra antinode 1/3 of the way from the open end and extra node 2/3 of the way. Now L is 3/4 of lambda, the n=2 case. This is the second harmonic. Adding more pairs of nodes and antinodes gives you further harmonics according to your formula.
    Note that the formula is different if both ends of the pipe are open.
    Frequency is speed of sound in the medium divided by wavelength: f = c/lambda.

    Check out http://www.physicsclassroom.com/Class/waves/u10l4c.cfm, http://www.physicsclassroom.com/Class/sound/u11l5d.cfm, http://www.physicsclassroom.com/Class/sound/u11l5c.cfm
     
  5. Apr 18, 2013 #4
    thanks
     
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