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Why fundermental mode of open-end string like that?

  1. Nov 14, 2008 #1

    KFC

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    I am so confusing why the fundamental mode of one-open-end string must satisfy that one-fourth wavelength equals to the length of the string. I read four textbooks but none of them tell why it is. And one textbook said we must make sure dy/dx=0 at the open-end, but I don't understand why.
     
  2. jcsd
  3. Nov 15, 2008 #2
    oh, I think I figured it out

    the free end of a string will always be an anti-node, and a wave is always flat at an anti-node, so we must always have the spatial derivative = 0 at the end (and at any antinode, for that matter)
     
  4. Nov 15, 2008 #3

    KFC

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    Thanks for your reply. I know the math, I just wonder what's physics origin of the requirement of the anti-node for an open-end string at the free end?
     
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