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Hooke's Law and Finding the Constant k

  1. Jul 15, 2007 #1
    Hooke's Law and Finding the Constant "k"

    1. The problem statement, all variables and given/known data
    What is the relationship between the extension in an elastic spring and the mass of an object suspended from it?

    2. Relevant equations
    F = mg
    F = -kx
    mg = -kx

    3. The attempt at a solution

    Here is my lab data:

    Now, in order to find the relationship, our teacher said we need to derive a relationship. But I don't know what to do. I am thinking that somehow I need to find the constant of the spring in order to find the relationship. The teacher also asked to use the equation to determine the extension when the mass is increased, for example, when the mass is 650 g.

    Also, here is my try to find the constant.
    mg = -kx
    (0.1)(-9.8) = -k (0.026)
    -0.98 / 0.026 = -k
    k = 37.69

    I am not sure what to do. Can anyone please help me? Thanks.
    Last edited: Jul 15, 2007
  2. jcsd
  3. Jul 15, 2007 #2

    You want to come up with a model that fits all of your data.

    Take note that F = mg (as you provided). Thus your step of (0.1)(9.8) was already done for you in one of your columns.

    So what happens if you graph the force (F) vs the displacement (x) ? In a linear relationship [itex] y = mx + b [/itex] what is [itex] m [/itex]?
  4. Jul 15, 2007 #3
    I graphed the Force vs. Extension. For "m" I got 18.85. And for "b" I am guessing that it must be 0.49 since that's the y-intercept.

    y = 18.85x + 0.49

    And I think this is pretty much it. Wow, I got the equation. But, I don't understand the relationship between the extension of the spring and the mass of an object.

    Please do help. Thanks.
  5. Jul 15, 2007 #4
    Guys, there is a question which asks me to find the extension when the mass 0.55 kg.

    Here are my calculations:

    F = mg
    = (5.5) (9.8)
    = 53.9 N

    y = 18.85x + 0.49
    53.9 = 18.85x + 0.49
    53.9 – 0.49 = 18.85x
    53.41 = 18.85x
    x = 53.41 / 18.85
    x = 2.83 m

    And then the question asks whether this is valid or not? Can someone please explain to me? Thanks.
  6. Jul 15, 2007 #5
    Anyone please help.
  7. Jul 15, 2007 #6
    Sorry man. I don't have the time right now. You could however Google something like "hookes law excel" or something like that. This is basically a standard lab for a physics course. You should be able to come up with something.

    Or you could wait for someone to answer. I'll be around later, so I can get back to you then. If you need faster help, either wait for someone else, or go with Google.
  8. Jul 15, 2007 #7
    Arite.Thanks for your help man. I will keep checking back to this topic and I am also trying to Google some topics on this activity. The bad part is that our teacher said we are not learning harmonics, which means that more than half the stuff I find is not any use to me. Thanks anyways. I will keep a watch on this topic.
  9. Jul 15, 2007 #8

    Ok. So you do no actually need to worry about the 0.49. Hooke's law [itex] F = -k x [/itex] implies a linear relationship. So what this means is if you pick a displacement value, lets say [itex] x = 2(cm) [/itex] the restoring force (it trying to pull back to the equilibrium position) [itex] F [/itex] is equal to the displacement [itex] x [/itex] multiplied by some constant [itex] k [/itex]. So if we pull back [itex] 2(cm) [/itex] the force would be [itex] F = (-k)(2cm) [/itex].

    So what you did was graph your data, and then find the constant (we can call it k, or c, or whatever) that when you move some value ([itex] x [/itex]), then what is the constant ([itex] m [/itex]) that when you multiply it by [itex] x [/itex] is equal to the Force. So you don't need the y-intercept, because you need to find the linear relationship. If you want me to be more mathematical let me know.

    You are actually dealing with harmonics! You are dealing with a simple harmonic oscillator. What this means is this. When you pull the weight on the spring down and let it go, the weight will shoot back up and go past is equilibrium point. It will then fall back down and then shoot back up. Well if you neglect friction and outside forces, the spring will continue in this motion. If you plotted the displacement from the equilibrium position, it would be that of a sine wave. This means it is a simple harmonic oscillator. There are other systems that have the same behavior, this is why the term is important.
  10. Jul 15, 2007 #9
    Thanks for your help.

    I drew the graph in Microsoft Excel, here is the picture:

    It would be helpful for me to see some mathematical calculations because it would be more practical. I would really appreciate that.

    I am a bit confused. Since I found out that the slope is 18.85, that came to be wrong. As I moved up, meaning that when I found the slope by selecting different points on the line, the slope came to be about 16.7. This is confusing because I thought the slope was the constant, and it was supposed to be the same throughout any set of points on the line.

    According to Microsoft Excel, when I used the linear trend line, the equation came to be y= 0.49x. But I am not sure whether that's right.

    Please help. Thanks.
  11. Jul 15, 2007 #10
    I tried to figure out the constant for each value of x (extension). I used the formula F = kx

    But to find k, I did : k = F / x


    For the first value, the extension is 0, so I didn't know what to put. So after that, I did the calculation for example:

    k = 0.98 / 0.026
    = 37.69 N/m

    I am still unsure what to do with this data since I don't fully understand the relationship between mass and the extension. Please help me more. Thanks.
  12. Jul 15, 2007 #11
    When you used excel to do the linear regression you were given an equation of the form:

    [tex] y = mx + b [/tex]

    What does this really mean? Well Excel found a function for you (using linear regression). This function is of the form [itex] F(x) [/itex] i.e. you give it a displacement (x) and it returns a force (F). This function is an approximation right?

    What happens when you plug in the displacement x=0.000001 ? Well guess what you never measured anything for that! So you really cannot say what the force is at that distance. Instead you now have a model that "fills in the blanks"; the model interprets what is between these measured data points.

    Now Hooke's law says that F=-kx, or that some constant (k) exists that when you multiply it by the displacement, it is equal to the force.

    Now you probably remember that a line of the form [itex] y = mx + b [/itex] that [itex] m = \frac{y_2 - y_1 }{x_2 - x_1} [/itex]. If you do this for two measured points, then that is the rate of change between those two points. However, that is NOT necessarily the rate of change for all the points. Your measurements when plotted may not necessarily all fit on a straight line. They will however be near a straight line. This is what hookes law gives you. It says that the measurements of a spring will all be close to some straight line. So when you plot your data in excel, and then use linear regression, you get a line that gets as close to all these data points as possible. Now some constant multiplied by the displacment will equal the force. Differentiate to find the rate of change [itex] \frac{d}{dx} (mx + b) = m [/itex]. This equals [itex] k [/itex]

    What is the regression line that excel gave?
  13. Jul 15, 2007 #12
    Actually, the regression line that excel gave was exactly on top of the line in the following graph:


    When I selected the option "Display equation on Chart" then it put up y = 0.49x.

    As for the constant, the formula that you put is really confusing to me.
    [itex] \frac{d}{dx} (mx + b) = m [/itex]
    Can you please explain this a little bit and how I may use this to find the constant?

  14. Jul 15, 2007 #13
    It is only y = 0.49x ?

    Have you taken calculus?

    How are you finding the slope?
  15. Jul 15, 2007 #14
    Ya, that's what excel gives me.

    Sorry, I haven't taken Calculus yet. But I have taken it for next year.
  16. Jul 15, 2007 #15
    Just disregard [itex] \frac{d}{dx} (mx + b) = m [/itex] for now. It is not hard to explain, it just requires an explanation that is unnecessary for this problem. I added some more questions on the post (via an edit).... if you wouldn't mind answering them, that would be cool.
  17. Jul 15, 2007 #16
    To find slope, (by using paper and pencil, not Excel), I drew the graph. Since the points were not exactly in a straight line (but almost close), I drew the line of best fit. Then, I took two points on the line.

    m = 1.8 - 1.3 / 0.08 - 0.05
    = 16.7 N /m.

    That's how I got the slope/constant. But I don't know why Excel is giving me y=0.49x. The Excel formula is right to one extent because if I put x = 2 then the second value of F is given, which is 0.98.

    But as for my calcuation, the slope is 16.7. Could you please tell me what to do next? Thanks.
  18. Jul 15, 2007 #17
    Using the extension and force data in this table, I used excel and got a regression line of:
    y=15.49x + 0.576
  19. Jul 15, 2007 #18
    Wow, that's amazing. If you could please follow through the steps and tell me how you got that equation, it would be helpful for me because I tried and I think maybe I made a mistake.
    Also, you used only the Force vs. Extension for the regression line right?
    Also, for the extension when it equals 0, I keep getting error now for some reason. I would like to know how you got the equation.

    Last edited: Jul 15, 2007
  20. Jul 15, 2007 #19
    I put in in column 1

    and column 2 had

    I then went to insert chart (scatter), and plotted it. I then clicked on the data and went to add trendline, and put the equation on the chart.

    You are trying to find a function that you can plug in an extension (x) (so this goes on the x-axis) that gives you a force (F) (this goes on the y-axis).

    Now k is 15.49 of that data (of course use units !)
  21. Jul 15, 2007 #20
    OMG! Thank you so much. All this time, I was using line graphs and that's the reason why I kept thinking that all the data was in a straight line.

    But now that you showed me to draw the scatter graph, I can clearly see the data is not straight just as it was on my graph that I drew by hand. Wow. I am so mad at myself.

    Thank you so much for explaining this to me. My only last question would be that the teacher has a question which asks that if mass is 5.5 kg, then what is the extension?

    So I use the equation:
    y = 15.49x + 0.576

    y = mg
    = 5.5 X 9.8
    = 53.9 N

    y = 15.49x + 0.576
    53.9 - 0.576 = 15.49x
    53.324 = 15.49x
    x = 3.44 m

    Now I know that 3. 44 m cannot be a valid value. But I have to explain it in terms so that teacher knows I am thinking.

    I found this website: http://www.darvill.clara.net/enforcemot/springs.htm

    It talks about permanent extension. But our teacher never taught us that. So would that be ok?

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