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Mass attached to lower of 2 identical springs

  1. Sep 22, 2013 #1
    1. The problem statement, all variables and given/known data
    a mass m hangs from a uniform spring of spring constant k.
    (a)what is the period of oscillations in the system?
    (b)what would it be if the mass m were hung so that:
    1)it was attached to two identical springs hanging side by side?
    2)it was attached to the lower of two identical springs connected end to end?

    (P.S. Im not sure if we are to neglect the force of weight of the mass)

    2. Relevant equations
    [itex]T=2\pi /\omega [/itex]

    3. The attempt at a solution
    (a)[itex]m\ddot{x} +kx=0[/itex]

    dividing by "m" you get: [itex]\ddot{x}+\omega^2 x=0[/itex]

    to find the roots: [itex]r^2 +\omega^2 = 0[/itex]
    to which the roots are: [itex]r=\pm i\omega[/itex]. According to this equation the angular frequency is [itex]\omega[/itex] which equals [itex]\sqrt{k/m}[/itex].

    Therefore, [itex]T=2\pi \sqrt{m/k}[/itex]

    (b)(1)skipping a few steps because its similar to last equation you get [itex]\ddot{x} +\omega^2 x=0[/itex]

    again finding the roots you get: [itex]\pm \omega \sqrt{2} i [/itex],

    According to this equation the angular frequency is [itex]\sqrt{2}\omega[/itex] which equals [itex]\sqrt{2k/m}[/itex].

    the period is: [itex]2\pi \sqrt{m/2k}[/itex]

    (b)(2) im not really sure how to set this one up but my thinking was something like this.
    I can write the spring force as if it were 1 spring of length 2L.

    but the thing is that the oscillations dont depend on the length of the spring, and im not sure if this is a correct way of going about it. Please help.
     
  2. jcsd
  3. Sep 22, 2013 #2
    For part 2, you need to figure out the stiffness of the spring combo. You could examine the static case (equilibrium) and draw the FBD.
     
  4. Sep 22, 2013 #3
    Right i did that, however i dont know how to combine the spring constants "k".
     
  5. Sep 22, 2013 #4
    Consider that a mass is attached to two springs (one on top of the other) of equal stiffness and the system is in equilibrium.

    What is the force acting on the top spring? On the bottom spring? How much extension does each spring have? What is the total extension? What is the total stiffness?
     
  6. Sep 22, 2013 #5
    in this "s" means the unstretched length of the spring, and "1" refers to top spring, "2" refers to bottom spring.

    F1=mg-k(s+x)

    F2=mg-k(s+x)
    where s is the natural length of spring

    spring 1 extends (s+x)
    spring 2 extends (s+x) as well, only from the bottom of the top spring.
    total extension=2(s+x)
    total stiffness=???
     
  7. Sep 22, 2013 #6
    Hooke's law is F = kx, not F = k(s + x). Drop the s'es in your equations.

    Other than that, you are almost there. Treat the two springs as one spring. You know the total force it is loaded with. You have computed its total extension. What is its stiffness then?
     
  8. Sep 23, 2013 #7
    After a mass m is attached to a spring, it stretches the spring by an amount s and attains a position of equilibrium at which its weight mg is balanced by the restoring force ks. which results in the final equation F=-kx. This way of looking at spring equations helps me understand basic equations better because when i am summing the forces, mg is an obvious one, and i need to know how to handle it.

    I want to say the stiffness would just be k because if you take a spring of stiffness k, and divide it into many small springs, i believe all those little springs would have stiffness k as well. But according to my textbook that is incorrect and i dont know why. I believe my textbook says it should be k/2.
     
  9. Sep 23, 2013 #8
    I figured it out. i didnt know how to take into account that only 1 spring was touching the mass. F=-k(x1+x2), F1=-k1x1, F2=-k2x2, set them equal and solve for x2, plug into F=-k(x1+x2), and you get 1/k=1/k1+1/k2
     
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