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Resonance and understanding what a resonant is?

  1. Oct 21, 2007 #1
    1. The problem statement, all variables and given/known data
    http://i215.photobucket.com/albums/cc10/Spookie71/image0-11.jpg

     
  2. jcsd
  3. Oct 21, 2007 #2
    Okay, you understand what it means for two waves to be superimposed (added together) right? Resonant waves are just one possible result when you add two waves together. Here's an applet that can let you see what happens with different frequencies:
    http://www.ngsir.netfirms.com/englishhtm/StatWave.htm

    If you max out the amplitude on the applet and set it to a frequency of 75 you can see a resonating wave with 4 nodes. At 99 you can see one with 5 nodes.

    Even though there are is one particular string, you can have different resonances (one with just two nodes, one with three nodes, 4 nodes... and so on.
     
  4. Oct 21, 2007 #3
    So those pictures that i posted for you are based on using different frequencies, that makes more sense to me. the applet makes alot more sense. The page didn't really say it was using different frequencies, just that there was a 2nd and 3rd Resonant.

    Still confused but the overall picture is becoming clearer.

    Thank you
     
  5. Oct 21, 2007 #4
    Can I help you with whatever you're still confused about?
     
  6. Oct 21, 2007 #5
    Anadyne I appreciate your help. to be honest I'm just a little overwhelmed. It's alot of new concepts all at once. Just a matter of processing them. I guess it would help a little if I knew where we are going and where I will apply this later in Physics. It's alot to remember how sound resonates in an air column that is open ended or closed on one end. Must sound silly to you but I'm having difficulty. Wish I could be more specific.

    Scott
     
  7. Oct 21, 2007 #6
    Ok here is a question for you, How can a column of air be closed at both ends? as their describing in this textbook?
     
  8. Oct 21, 2007 #7
    Actually forget my last question:

    I have a question regarding an equation again.

    Required is [tex]\lambda[/tex]

    l = [tex]\frac{(2n - 1)\lambda}{4}[/tex]

    becomes

    [tex]\lambda[/tex][tex]\frac{4l}{(2n - 1)}[/tex]

    I understand that you multiply the 4 by both sides and then bring the 4l over as the numerator. Where it becomes confusing is the (2n -1) and bringing it down to the denomenator. Doesn't bringing it down from the numerator cancel it out?
     
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