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    1st Harmonic - 20Hz - F0 -wavelength 2L
    2nd Harmonic - 40Hz - 2F0 -wavelength L
    3rd Harmonic - 60Hz - 3F0 - wavelength 2/3L

    Can someone explain this how does the frequency relate to the wavelength as the harmonic increases?

    If the Length of the string is L how does frequency effect this? How do you work out the wavelength?
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    (Original post by Acrux)
    1st Harmonic - 20Hz - F0 -wavelength 2L
    2nd Harmonic - 40Hz - 2F0 -wavelength L
    3rd Harmonic - 60Hz - 3F0 - wavelength 2/3L

    Can someone explain this how does the frequency relate to the wavelength as the harmonic increases?

    If the Length of the string is L how does frequency effect this? How do you work out the wavelength?
    The equation to remember is:

    wavelength / 2 = L

    So if we have 3 fll wavlengths in the tube (standing waves), then it becomes:

    3wavelengths / 2 = L
    Rearrange to get wavlength:

    Wavelength = 2L/3

    HIgher harmonic means more wavelengths in the same tube/length, so the frequence will increase as there are more per unit time/length.

    So: wavelength decreases as frequence increases.
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    (Original post by derpz)
    The equation to remember is:

    wavelength / 2 = L

    So if we have 3 fll wavlengths in the tube (standing waves), then it becomes:

    3wavelengths / 2 = L
    Rearrange to get wavlength:

    Wavelength = 2L/3

    HIgher harmonic means more wavelengths in the same tube/length, so the frequence will increase as there are more per unit time/length.

    So: wavelength decreases as frequence increases.
    Is this the seperation between the adjacent nodes?

    Secondly how do workout how many waves there are? So for the 3rd harmonic there is 3 wavelengths?
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    (Original post by Acrux)
    Is this the seperation between the adjacent nodes?

    Secondly how do workout how many waves there are? So for the 3rd harmonic there is 3 wavelengths?
    The seperation is between adjacent nodes or antinodes.

    For the second question, do you mean how many ful wavenlengths, because there is 1 standing wave, but 1.5 wavlengths. Im assuming you got 3 which is the number of full circles? Remember that one full circle is only half a wavelength, so 3 full circles = 3/2 wavelengths.
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    (Original post by derpz)
    The seperation is between adjacent nodes or antinodes.

    For the second question, do you mean how many ful wavenlengths, because there is 1 standing wave, but 1.5 wavlengths. Im assuming you got 3 which is the number of full circles? Remember that one full circle is only half a wavelength, so 3 full circles = 3/2 wavelengths.
    Yes i mean full wavelength how are you getting 3 wavelengths? Yh i dont understand the working behind that
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    (Original post by Acrux)
    Yes i mean full wavelength how are you getting 3 wavelengths? Yh i dont understand the working behind that
    Think I've confused myself, which harmonic are we looking at?
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    (Original post by derpz)
    Think I've confused myself, which harmonic are we looking at?
    3rd harmonic has 4nodes and 3 Antinodes
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    (Original post by Acrux)
    3rd harmonic has 4nodes and 3 Antinodes
    The way I do it, I just look at the whole number of circles which in this case is 3 and divide it by 2 since each one it half a wavlength, sorry about the confusion
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    (Original post by derpz)
    The way I do it, I just look at the whole number of circles which in this case is 3 and divide it by 2 since each one it half a wavlength, sorry about the confusion
    3rd harmonic has 4nodes and 3 Antinodes
    So one wavelength is 3/2 right

    For the 4th harmonic one wavelength is 1/2

    but for the 5th how many is one wavelength? How do you work this out
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    (Original post by Acrux)
    3rd harmonic has 4nodes and 3 Antinodes
    So one wavelength is 3/2 right

    For the 4th harmonic one wavelength is 1/2

    but for the 5th how many is one wavelength? How do you work this out
    I thought you only need to know up to the 3rd harmonic. Also, I think it depends on if its an open ended or close ended tube
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    (Original post by derpz)
    I thought you only need to know up to the 3rd harmonic. Also, I think it depends on if its an open ended or close ended tube

    well its say the L is 2/5L
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    (Original post by Acrux)
    well its say the L is 2/5L
    So λ= 2/5L

    Therefore
    /2 = L
    So the wavelength is 2.5
 
 
 
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