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Homework Help: The speed of sound

  1. Apr 17, 2007 #1
    Alright we did a lab using the speed of sound we were suppose to able to calculate the temperature outside.
    v = 332 +0.6T

    So this is what we did, one person hit two blocks together twenty times at an equal pace. The timers start when the person first hits the block together and stops when the last echo is heard.
    The distance from the wall and back is 200m
    So the times we got were 23.40s, 23.28s

    Then I calculated the average total time and average time for each echo cycle. Which is 1.167s

    So Speed = Distance / Time for each cycle Right?
    Speed = 200m/1.167s?
    If I do this, I end up with a negative temperature.

    So then did this, Speed = Distance/ Half of the time for each cycle.
    Speed = 200m/0.5835s
    By doing this I got the right temperature.

    Can anyone explain to me why I needed to divide the time for each cycle by two?
  2. jcsd
  3. Apr 17, 2007 #2


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    If your time is from when you clap to when you hear the echo, the distance is actually 400m since the sound is making a round trip. (Or half the total time to go 200 m)
  4. Apr 17, 2007 #3
    What do you mean its making a round trip?
    Do you mean like its 400m to the wall and back?
    Because the distance to the wall is 100m.
  5. Apr 17, 2007 #4


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    Hmmm, I misread part of your post, I thought the 200 m was the distance to the wall only. Sorry about that.
  6. Apr 17, 2007 #5
    Yea its no problem. But do you understand why the time would be divided by two or why the distance would be multiplied by two?
  7. Apr 17, 2007 #6
    helluva way to measure temperature if you ask me:tongue:

    But something is amiss, 200/1.17=170m/s, exactly 1/2 the speed of sound. I think maybe something in the methods used which caused the receiver to hear an earlier echo than the one triggered by the last event? I would think cleaner, tho less efficient and I understand the reasoning I believe, but led to big error anyway, single trials x 20..
  8. Apr 17, 2007 #7
    Huh? =/ lol sorry, I still dont understand why its like this.
  9. Apr 17, 2007 #8


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    Just to check, are you SURE the distance to the wall and back was 200 m??

    I think what denverdoc is saying is that it could be due to error in measurement, like stopping the timer too soon, or maybe just in the reaction time of the person working the timer.
  10. Apr 17, 2007 #9
    The notion I suppose is a training effect and cancels sone randomness from the human side, in other words observations 10-20 show less
    variability than 1-10. Thats the strength of the experiment. The weakness is hearing slap/echo/slap/echo and avoiding echo/slap error. In plain english after ten trials are the timekepers hearing the sound fom an earlier echo?
    Last edited: Apr 17, 2007
  11. Apr 17, 2007 #10
    Yup I'm sure the distance there and back is 200m.
  12. Apr 17, 2007 #11
    No I don't think the time keepers are hearing the sound from an earlier echo, it was pretty clear 20 echoes were heard.
  13. Apr 18, 2007 #12
    i didn't buy it either. especially if it was just an aggregate time. Hage is right, had to have been 50 meters away. 100m is like a football field.
  14. Apr 18, 2007 #13
    alright I found out why it is now, by hitting it at an equal pace, there is a pause between each echo and the next hit.

    Hit-(speed moves)-Echo-(pause to create equal pace)-Hit-(speed moves)-Echo-(pause to create equal pace)

    So the 1.167s is the time between the hits. Therefore I had to divide it by two.
  15. Apr 18, 2007 #14
    good grief, why didn't i see that, thanks for the update.
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