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Plot question

  1. Feb 28, 2010 #1
    I have to plot the average period squared vs the mass, now my question is which goes to the x axis and which goes to the y?
     
  2. jcsd
  3. Feb 28, 2010 #2
    X: Independent variable
    Y: Dependent variable.

    So, did you change mass to find different periods or did you change the periods of find different masses?
     
  4. Feb 28, 2010 #3
    i added mass to see the variation in oscillation periods
     
  5. Feb 28, 2010 #4
    That means that the period DEPENDS on the mass, making mass the independent variable and period the dependent variable.

    Remember:
    X is what you change
    Y is what you measure
     
  6. Feb 28, 2010 #5
    for example to calculate the k constant i believe the formula is

    k = (4pi²)/slope

    but the slope of the graph has to be t² and kilograms correct? i have the values in t² and the mass in grams
     
  7. Feb 28, 2010 #6
    If you want the spring constant to come out as Newtons per meter then yes the mass has to be in kilograms.
     
  8. Feb 28, 2010 #7
    Is it supposed to give a huge difference from the elongation method?

    My assignment says i have to do 2 tables where one measures force against displacement and than this one above

    for force vs displacement i get k as 34.6 while when i do it with the formula i got above i get k as-8.77, is this normal?
     
  9. Feb 28, 2010 #8
    You can't calculate the slope the same way for force vs. displacement. Remember, [tex]F = kx[/tex], so when you graph that equation you are going to end up with a line anyways. What do you notice about the slope?
     
  10. Feb 28, 2010 #9
    The instructions say that the slope of the graph force vs displacement which i got a slope of 34.6 N/m and when i did the slope calculation of the oscillation graph i got -4.5 for the slope where i imputed in the formula

    k = (4pi²)/slope

    which gave me -8.77 N/m
     
  11. Feb 28, 2010 #10
    If you plot displacement against weight, yes. If you plot it against mass, the k = m * 9.8
     
  12. Feb 28, 2010 #11
    The first graph i have
    y axis - Force (N)
    x axis - displacement (m)

    Where i got the slope of 34.6 which the instructions say its the k constant

    the second graph
    y axis - t²
    x axis - mass (kg)

    where i got slope of -4.5

    which i imputed here

    k = (4pi²)/slope

    which gave me -8.77 N/m

    Now i was supposed to calculate the difference between the two

    with these values i get a difference of -125% which i think is a huge difference
     
  13. Feb 28, 2010 #12
    You swapped X and Y on the first graph. You change force and measure displacement.

    I'm not quite sure how you ended up with a negative slope, there should be a strong positive correlation between mass and period squared. Besides, the spring constant can't be negative. Can you list some data points?
     
  14. Feb 28, 2010 #13
    First graph

    x f(N)
    0.98
    1.96
    2.94
    3.92

    y displacement (m)
    0.03
    0.06
    0.088
    0.115


    Second graph

    x mass(kg)
    .3
    .4
    .5
    .6

    y t²(s)
    2.72
    2.13
    1.66
    1.41

    i used this website to help input data to get info
    http://www.shodor.com/unchem/math/lls/leastsq.html
     
    Last edited: Feb 28, 2010
  15. Feb 28, 2010 #14
    Ok the first graph seems fine but I can't understand why there's a negative slope on the second one. Can you make sure you're using periods and not frequencies? That's the only thing I can think of that would decrease like that.
     
  16. Feb 28, 2010 #15
    we had to count how many secs it took to do 25 oscillations

    .3kg - took 15.19 sec
    .4kg - 17.16 sec
    .5kg - 19.38 sec
    .6kg- 21.09 sec

    than i took the average period T (sec)

    divided 25 (oscillations) by the secs it took and squared them

    25/15.19 = 1.65 = t² = 2.72
    25/17.16 = 1.46 = t² = 2.13
    25/19.38 = 1.29 = t² = 1.66
    25/21.09 = 1.19 = t² = 1.41


    and thats how i got my values.

    in a way the graph makes sense, the higher the mass the less time it takes to do an oscillation but i know o have something wrong on this graph
     
    Last edited: Feb 28, 2010
  17. Feb 28, 2010 #16
    Ah, that's it. An oscillation divided by time is a frequency. You want to use the reciprocal of that (period). I did that for the excel chart I made of your data and got both measurements of k almost identical.
     
  18. Mar 1, 2010 #17
    So i am supposed to divide these values by one and than square them, is that what i did wrong?

    25/15.19 = 1.65 = t² = 2.72
    25/17.16 = 1.46 = t² = 2.13
    25/19.38 = 1.29 = t² = 1.66
    25/21.09 = 1.19 = t² = 1.41
     
    Last edited: Mar 1, 2010
  19. Mar 1, 2010 #18
    Or just divide the values you already have by one since you already squared them. If you don't understand why, frequency is how many oscillations, revolutions, etc. the object makes per second, thus having the unit of hertz. Period is how many seconds it takes the object to make one revolution.

    The frequency of the earth is 1/24 hours = 0.04 per hour (meaning that it makes 1/24 of a rotation per hour) and its period is 24 hours (meaning that it takes 24 hours to make one revolution.
     
  20. Mar 1, 2010 #19
    Your a life saver, i have been around for so long that i was getting mad lol I just have 1 more question i had to calculate the energy stored in one of the spring, the units of the energy stored is in Joules correct?
     
  21. Mar 1, 2010 #20
    Yes, so long as the spring constant is N/m and the displacement is m.
     
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