1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Double spring question

  1. Oct 25, 2007 #1
    1. The problem statement, all variables and given/known data
    A block of mass 20kg with a spring with a stiffness of 500N/m attached to the bottom is dropped .5m from rest onto another spring of stiffness 800N/m. The weight of the springs can be neglected. What is the maximum deformation of each spring due to the collision.

    2. Relevant equations
    work = change in kinetic energy + change in potential energy + change in elastic energy

    3. The attempt at a solution
    I have assumed that I can add the different spring coefficients, giving me an energy equation of:

    work=0= -.5mv^2 + -mg(.5+x) + .5k(x^2)

    here k is the sum of the spring coefficents and x is the total deformation of the spring.
    I solved the equation, got two answers for x one positive and one negative.

    Can I use the ratio of the spring coefficients to work out the individual reformation, and also do i use the positive value calculated for x?

    Sorry for the long question, your help is appreciated.
    Last edited: Oct 25, 2007
  2. jcsd
  3. Oct 25, 2007 #2
    Ermm, if the springs are inline with each other (on top of each other) you add the reciprocals of the constants, so that:

    [tex]{k_T}^{-1} = {k_1}^{-1} + {k_2}^{-1}[/tex]

    The system adds (becomes stiffer) when the springs are used adjacent to each other.

    So I think, from the wording of the question, you need to use this equation maybe:

    [tex]k_T = \frac{k_1 k_2}{k_1 + k_2}[/tex] - Your thoughts

    But yeah - the kinetic energy term would become [itex]mg\Delta h[/itex] in your equation (because the kinetic energy came about from falling the distance [itex]\Delta h[/itex], cause when it hits the spring, your [itex]mgh = 0[/itex]. Apart from that, logic seems fine...

    Let me know how you get on :P
  4. Oct 25, 2007 #3
    I do agree with Sam.. Keq = k.k'/(k + k') , is the one to be used.
    Since, the two springs are adjacent, force in them must be equal => k.x = k'.x'
    Equivalent spring, Keq.X = k.x = k'.x', where X = x + x'.

    @ sumeer@dinsum.c
    I posted this, just to urge one thing: please REASON OUT your assumptions! Do not make blind assumptions. The assumptions you made (that Keq = k+ k') would have been true, if elongations in both the springs would have been same! x = x', X = x = x', and Ftot = Keq.X = k.x + k'.x'
    I hope you get the idea.
  5. Oct 25, 2007 #4
    Hi Sam,

    Thanks for your information, I've managed to work through the problem.
    Last edited: Oct 25, 2007
  6. Oct 26, 2007 #5


    User Avatar
    Homework Helper

    Hi Sameer, I don't how you solved the problem. But see whether it matches with this. Loss of PE of block = mg(0.5 +X). This energy is shared by the two springs and it is equal to 1/2*kx^2 + 1/2*k'x'^2. During the compression force in each spring is kx and k'x'. Hence down ward force =mg + kx and upward force = k'x'. In equilibrium position mg +kx = k'x' and X = x+x', write x' and X in terms of x and substitute it in the above energy equation. Then solve for x, and x'
    Last edited: Oct 26, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Double spring question
  1. Springs Double (Replies: 3)

  2. Spring Questions (Replies: 1)