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Homework Help: Hooke's Law Practical activity

  1. Apr 19, 2007 #1
    1. The problem statement, all variables and given/known data

    This isn't acctually a problem but I really don't know what to do. I was doing a simple activity about Hooke's Law during class today where I have to set up a apparatus and measure the extension of spring when when we put differnt masses on it. I didn't do it properly so I have difficulty answering it

    This is my graph if it helps any

    http://img156.imageshack.us/img156/8913/clipuw9.jpg [Broken]
    you see, I only did the experiment with 1 spring. I didn't do the following required experiments, so I have no idea how to answer.
    2. Relevant equations

    Q6. Write how much extension do you think there would be (compared to the extension for one spring) if you now had two springs, identical to the one used, joined one after the other

    Q7. Observe what happens (describe). Explain

    Q8. Do the same, but this time for two springs side by side

    Q9 What about three side by side, or four, or three in a row, or four?

    3. The attempt at a solution

    Q6. I think it remains the same cuz F and k remains the same => x remains
    Q7. Observation would suit what I just said
    Q8. 2 springs side by side so forced is distributed even onto both springs thus making the extension halved
    Q9. I don't even understand the question

    Thanks in advance:!!)
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Apr 19, 2007 #2
    I like q8 response a lot!

    So lets call a new spring constant K'eq that which satisfies the eqn,

    f=-kx. the force is the same, and as you propose extention is halved,
    f=-keq*.5x divide be
    f=-kx and you will find k'eq=2k. For q9 extend your logic for 3 or more springs.

    now q6 is a bit more subtle, but if you think more about it, you may start to doubt your answer.
  4. Apr 20, 2007 #3
    I don't get why Q6 response is wrong, well as far as I understand force constant of a spring shows us how stiff it is. If two identical springs connected in 'series', the stiffness of the 'combined' spring doesn't change. If we don't increase the mass, i.e. no change in weight force, the extension (x) of the spring shouldn't change.
  5. Apr 22, 2007 #4
    err, no suggestion? T_T I keep on think my way of thinking is right and some online experiment tells me that it's wrong. Explanation please
  6. Apr 22, 2007 #5
    Been gone, think about this thought experiment forget on-line proof etc. You have a bungee like cord and want to stretch it by 2 inches. Cut the bungee cord in half and see how much more effort it requires to stretch the same 2 inches. Or if you want to save yourself a buck and have some rubberbands handy, for grins tie 4 in series and compare that with one in terms of stretch force and displacement.
    The math is relatively trivial thing to prove, but I'm thinking it might be more helpful to get it out of the purely math realm. Hope this helps a bit.
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