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Quick Question ( y = mx + c )

  1. Feb 26, 2008 #1
    [SOLVED] Quick Question ( y = mx + c )

    Okay, this shouldn't take long. Would it be okay if I did not show any working as I want to know how one equation related to another, which will then aid me in trying to solve the actual question. [tex]T^2[/tex] is the y-axis and [tex]d^2[/tex] is along the x-axis.

    The equation of the straight line I have drawn is...


    I am asked to find a value for for K, and then to calculate a value for m.

    m = the effecitve mass
    K = the constant for the system

    Now I would have thought I would need to consturct a similtaneous equation, but I have a feeling that this equation can be linked to...


    Would x and y in this equation relate to my axis? What would K be? If I put the intial equation in the form [tex]y=mx+c[/tex] will it work? If so here is how I would have thought it should look.

    [tex]T^2=m\times d^2 + c[/tex]

    Now I can put in the values I have for [tex]T^2[/tex] and [tex]d^2[/tex] so...

    [tex]3.13=m\times 0.026 + c[/tex]

    I am still not sure what c is....is it K?

    Any help would be great. If this does not make sense please ask for anything that isn't there and I will go and find it :tongue:
  2. jcsd
  3. Feb 26, 2008 #2
    If you know T2 and d2, you can rewrite the first equation you wrote as:

    K = T2 / (0.10d2 + 0.042m)

    This equation has the form y = a / (b + x), not y = mx + b.

    This assumes that K depends on m.
  4. Feb 26, 2008 #3
    The problem is that I do not know the values for m or k, so I cannot calculate either of them from that.
  5. Feb 26, 2008 #4


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    You can calculate K from the gradient. Your equation is [itex]T^2=(0.1K)d^2+0.042Km[/itex] which is, on letting y=T^2 and x=d^2 in the form [itex]y=nx+c[/itex] where n=0.1K and c=0.042Km. Now, what does n represent? Can you obtain this from the graph? If so, then you have K and can use your y intercept to find m.
  6. Feb 26, 2008 #5
    What is n? n isn't in any of my calculations? I know what y and x represent but I do not know what y stands for. m is the gradient...
  7. Feb 26, 2008 #6


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    I changed n to be the gradient when I wrote the equation y=nx+c. You've got two m's and so that's bound to be confusing. Anyway, the important point is that you have the equation in the form y=(gradient)x + (intercept). Compare this to [itex]T^2=(0.1K)d^2+0.042Km,[/itex] with y=T^2 and x=d^2. What is the gradient of your graph? What is the y-intercept?
  8. Feb 26, 2008 #7
    [tex]Gradient = \frac{T^2}{d^2}[/tex]

    I calculated this at 3.3, I did this by drawing a triangle onto the graph to determine this value.
  9. Feb 26, 2008 #8


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    Ok good, but I think you mean gradient= delta(T^2)/delta(d^2). Now you have that the gradient=3.3=0.1K. From this you should be able to find K.
  10. Feb 26, 2008 #9
    Thanks for your time Cristo, much appreciated. :smile:
  11. Feb 26, 2008 #10


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    You're welcome!!
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