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Standing Waves in a closed pipe

  1. Apr 6, 2010 #1
    So confused about standing waves in a closed pipe, which is open at one end and closed at the other. The closed end has a node while the open end has an antinode. To figure the wavelength, i use the formula:
    Lambda = 4L/n where n is the number of harmonic and can only be odd integers
    frequency of the wave, f = nv/4L

    How can i figure out the number of nodes or antinodes? For example, the first harmonic is the fundamental frequency and has one node and one antinode. the third harmonic is the first overtone and has two nodes and two nodes. the fifth harmonic is the second overtone and has three nodes and three antinodes. What if I come across a situation where i'm dealing with n=53, is there a formula i can employ to figure out the number of nodes and antinodes?

  2. jcsd
  3. Apr 6, 2010 #2
    Draw the first few standing waves and generalize. You will find that the n-th harmonic has n+1 nodes and anti-nodes (combined). If n+1 is even there will be (n+1)/2 nodes and (n+1)/2 anti-nodes. If n+1 is odd, there will be (n+2)/2 nodes and n/2 anti-nodes.
  4. Apr 6, 2010 #3
    I tried your recommendation in a problem in my mcat book and it works great. For some reason, the book doesn't discuss how to figure out nodes and antinodes. Thank you so much, really appreciate it!!
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