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How to graph a linear relationship between T and v of sound?

  1. Apr 7, 2016 #1
    1. The problem statement, all variables and given/known data
    I did an experiment involving the speed of sound at different temperatures. I placed two microphones at a fixed distance apart and I measured the time taken for a sound wave to travel between the two mics. I repeated this for different temperatures. I want to make a graph for my results, but I want a straight line relationship.

    2. Relevant equations
    v=d/t , v = 331 + 0.6T

    3. The attempt at a solution
    I could manually work out speed using v=d/t and plot v against the Temperature, T. However I do not know if this will give me a straight line. Or I can also plot the time against the temperature.
    Extra question: in both cases, will the gradient be of any use to me?
     
  2. jcsd
  3. Apr 7, 2016 #2

    BvU

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    Why hesitate ? Make the plot and post if you have questions ! :smile:
     
  4. Apr 8, 2016 #3
    Alright, I plotted speed of sound against temperature in Kelvin. I tried to replicate the formula v = 331 +0.6T
    Mine went horribly wrong. I got v = -675 + 3.07T
     
  5. Apr 8, 2016 #4

    BvU

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    My telepathic capabilities are sufficient to suspect a mistake :smile: but to actually help out more information is necessary ...
     
  6. Apr 8, 2016 #5
    I calculated the speed of sound using v=d/t where d is the fixed distance between each of the two mics and t is the time taken for the sound wave to travel between them(fast timer was used). I plotted this and their respective temperatures(in K).
     
  7. Apr 8, 2016 #6

    BvU

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    So you measured 163 m/s at 0 ##^\circ##C, 224 at 20 ##^\circ##C and 286 at 40 ##^\circ##C ?
     
  8. Apr 8, 2016 #7
    I only measured the speed at five intervals between 25 and 40 degrees celsius. The line I found is Excel's best fit line, but yes, I recorded something similar to 286 for 40 degrees.
     
  9. Apr 8, 2016 #8

    Ray Vickson

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    Do you have any reason to suppose the relationship should be a straight line?
     
  10. Apr 8, 2016 #9
    I found the equation online(v = 331.4 +0.6T). That is similar to y=mx + c, the equation of a straight line. So I am going under the assumption that that is what my results should mirror.
     
  11. Apr 8, 2016 #10

    BvU

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    Yes, this is a common expression. But T is in ##^\circ##C, not Kelvin.
     
  12. Apr 8, 2016 #11
    Ahh. I misinterpreted a graph I saw online and thought I had to convert into K.
    I plot another graph and this time I got the result v = 164 + 3.07T
     
  13. Apr 8, 2016 #12

    BvU

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    Yes, the slope doesn't change if you add a constant to the abscissa.
     
  14. Apr 8, 2016 #13
    So how would I conclude my experiment? Because obviously my results don't fall in line with the standard. Is it scientifically accurate to call this experiment 'inaccurate'?
     
  15. Apr 8, 2016 #14

    BvU

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    You don't show your observations -- at least from an earlier post I understand that the 'similar to 286 m/s' is the outcome of a calculation -- so it's hard to comment. For you it will also be hard to backtrack to find out if something went wrong.

    In itself that is a very valuable learning experience: evaluate while doing the measurements so you can change things / do extra checks if results are not as expected.

    Since the speed of sound is expected to be well above 300 m/s, a result below needs scrutinizing. Distance between microphones ? Time difference ? Path of the sound ?
    It would be well worth it to place the microphones further apart to see if the calculated ##\Delta x\over \Delta t## is really independent of ##\Delta x##. If that isn''t the case the (much harder?) effort to vary the temperature isn't very useful.

    Hard to say without further info. If your distance measurement has an accuracy of 10% and your delta time measurement also, a deviation of the "litterature value" of 355 - 286 = 70 m/s is not exceptional, but if both are supposed to be about 1% you have a problem.
     
  16. Apr 8, 2016 #15
    I worked out the uncertainty in my gradient and y-intercept using the parallelogram method and my final answer is 164+ 3.07T, with the uncertainty in gradient being plus or minus 1.30 and the uncertainty in the y-intercept being plus or minus 44.8. So the closest I could get to the actual equation is 208.8 + 1.77T.
     
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