1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Speed of Sound

  1. Mar 4, 2006 #1
    Hello, I'm new here and I'm not sure wether this is the correct forum or not, ohwell here it goes.
    I'm wondering how to calculate the speed of sound by using the formula:
    Lamba = v * T
    The frequency was 480 Hz and the distance was 33.34 cm.

    This was done in a tube with sand so you could distinguish the soundwaves in the sand thus measuring the distance of one wave length.
    Any help would be appreciated!
     
  2. jcsd
  3. Mar 4, 2006 #2

    LeonhardEuler

    User Avatar
    Gold Member

    T here means the period, which is one over the frequency:
    [tex]T=\frac{1}{f}[/tex]
    [tex]f=\frac{1}{T}[/tex]
    So you can calculate the period from the frequency. You can solve for "v" in the equation you gave and use the values of T and [itex]\lambda[/itex] to get a number. Just remember that your [itex]\lambda[/itex] is in centimeters so you should convert if you want a speed in m/s.
     
  4. Mar 4, 2006 #3
    So basicly the speed is 0,3334 * 480 = 160.03 m/s ??
    Sorry, but I'm a bit slow and this isn't my primary language either.

    Another question, what if I wish to calculate the theoretical resonance wave length for the tube, (75 centimeters of lenght, closed in one end, open in the other end).
     
  5. Mar 4, 2006 #4

    LeonhardEuler

    User Avatar
    Gold Member

    Right.
    Since it is closed on one end, there will be a node at that end (a point of zero amplitude). Sice it is open at the other, there will be a maximum there. So there will be resonence when the lenth of the tube is 1/2[itex]\lambda[/itex], 3/2[itex]\lambda[/itex], 5/2[itex]\lambda[/itex]...
     
  6. Mar 4, 2006 #5
    Thank you for your answers so far, but how do I calculate the resonance? (if that's possible)
     
  7. Mar 4, 2006 #6

    LeonhardEuler

    User Avatar
    Gold Member

    There will be resonence when the length of the tube is (1/2)[itex]\lambda[/itex], (3/2)[itex]\lambda[/itex]... and so on. So the longest wavelength at which there will be resonence is when
    [tex]\frac{1}{2}\lambda=L[/tex]
    (L is the length of the tube). So you just have to solve for [itex]\lambda[/itex].
     
  8. Mar 4, 2006 #7
    Alright
    So I calculated it like this: (meters)
    L1 = 1/2 * 1.5
    L2 = 3/2 * 0.5
    L3 = 5/2 * 0.3
    And it checks out.
    Any comments?
    I'm also supposed to compare this wave length to the ones I noticed during my experiment, the biggest one I got was 0.3334 meters and the smallest one was 0.07 meters, what does this say?
    So here it would be the wave length of 1.5 meters as the biggest and 0.3 meters as the smallest. Why is it so much bigger than the wave length that I got?
     
    Last edited: Mar 4, 2006
  9. Mar 4, 2006 #8
    remember that f=1/T when doing this calculation
     
  10. Mar 4, 2006 #9

    LeonhardEuler

    User Avatar
    Gold Member

    He did. He moved it over to the other side.
     
  11. Mar 4, 2006 #10
    ooops sorry, its late and im tired, should probably get myself off to bed, just one more crack at my optics and i will i think
     
  12. Mar 4, 2006 #11
    It's okay, those calculations are fine now, although I'd appreciate some feedback on my previous post!
     
    Last edited: Mar 4, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Speed of Sound
  1. Speed of Sound (Replies: 1)

  2. Speed of sound (Replies: 2)

  3. Speed of sound (Replies: 2)

  4. Speed of Sound (Replies: 5)

  5. Speed of Sound? (Replies: 6)

Loading...