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Spring Constant given mass, and equivalent mass in free fall from fixed height.

  1. Apr 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Two 54kg blocks are held 30cm above a table. As shown in the figure, one of them is just touching a 30cm long spring. The blocks are released at the same time. The block on the left hits the table at exactly the same instant as the block on the right first comes to an instantaneous rest.

    What is the spring constant?

    2. Relevant equations
    Delta L = mg/k

    3. The attempt at a solution
    I think the solution might start with finding the time it takes for the free fall block to hit the surface, which is the same as the block mounted on the spring. Then we can use to some how find the compression length. But that's as far as I have got.
  2. jcsd
  3. Apr 18, 2010 #2
    Yes, right. First of all, you can calculate the time that the block without a spring in its belly needs in order to reach the floor and equate this with the time that the other block needs in order to first come to instantaneous rest. Of course the second block's motion is going to be a part of a simple harmonic oscillation, so the time until first instantaneous rest is a quarter of the oscillation's period, meaning [itex]t = T/4[/itex]. Use that equation to calculate the spring constant.
  4. Apr 18, 2010 #3
    Awesome, so this is what I did from your suggestion:

    d=vi * t + 0.5 * a * t^2
    0.3m = 0 + 0.5 * 9.8 * t^2
    t = 0.2473 seconds

    0.2573 * 4 = 0.9897 = period

    frequency = 1 / period
    frequency = 1.01 hz

    frequency = (1/2pie) * sqrt(k/mass)
    1.01 = (1/2pie) * sqrt(k/0.054kg)
    k = 2.176 N/m

    But however, my professor told me this was wrong... please let me know where I made mistakes.
  5. Apr 18, 2010 #4
    Oh yea, that's true, my bad. You see, the point of balance for the oscillation of the second block isn't the point that it initially is, where it touches the spring, but a little below that, so that the force from the spring is opposite to the block's weight. The second block first reaches that point and T/4 later it reaches it's first instantaneous rest point.

    There's a chance we're talking about different things here. When saying instantaneous rest point you mean i) the point where ΣF=0 for the second block or ii) the point where v=0 (velocity) for the second block?? Because I'm not perfectly familiar with the terms in english.
    Last edited: Apr 18, 2010
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