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A spring question

  1. Oct 26, 2008 #1
    I am rather confused about springs. This may be an obvious question, so just bear with me.

    Now, say you had two spheres of any mass, a spring (that obeys Hooke's Law) connecting them, and the whole system was without external forces or friction (we can say, suspended in space). Now, what would happen if you applied a force to one of the spheres, in the direction of the other sphere? How far would the spring contract - would it even contract in the first place?
  2. jcsd
  3. Oct 26, 2008 #2


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    No. If there is no opposing force the spring won't compress. There must be a reaction force on the opposite end of the spring for it to compress. Otherwise, it will just move in space.

  4. Oct 26, 2008 #3
    I disagree. Lets say we have two masses attached to a spring, m1 and m2. The spring constant is known and is equal to k. Let's say we push the masses with force F until the entire system is accelerating at a. We know, from free-body diagrams, that F-Fspring = m1*a and Fspring = m2*a. Eliminating a, we have (F - Fspring)/m1 = Fspring/m2. Solving for Fsrping we have (F*m2)/((m1+m2)) = Fspring = kx. Thus the spring will compress until x = (F*m2)/(k*(m1+m2))
  5. Oct 26, 2008 #4


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    Staff: Mentor

    The reaction force is due to the inertia of the spheres via f=ma. A force on one causes both to accelerate and a force between them (it'll be half the force applied to the first).
  6. Oct 27, 2008 #5


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    Opps! Forgot about inertia...you can go ahead and smack me now. :redface:

  7. Oct 27, 2008 #6
    You can decribe the response of the spheres with respect to each other with a simple second order DE.

    mx'' - kx = 0
  8. Nov 2, 2008 #7
    How would I apply this equation? Would I use it for both spheres?

    And if I wanted to add in damping, the equation would be this: mx'' - cx' - kx = 0 where c is the damping coefficient?

    Thank you all for your help.
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