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SHM Mass Hanging on a Spring - Find velocity at given position in path

  1. Aug 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Spring and Hanging Mass
    A block of mass 4 kg hangs vertically from a spring with constant k = 100 N/m which is attached to the ceiling and initially is not stretched or compressed. The block is initially held at rest by an external force. The block is then released and falls under the combined forces of gravity and the spring.

    What is the speed of the block when it has dropped a vertical distance 0.4 m?

    (Quick note: I defined down as positive and up as negative)

    2. Relevant equations
    F (spring) = -k*x
    F (weight) = m*g
    F (net) = F (spring) + F (weight) = m*a
    a = ω^2*x
    v = ω*r

    OR I could go this direction

    PE (elastic) = 1/2*k*x^2
    KE = 1/2*m*v^2
    PE = KE (by conservation of energy)

    3. The attempt at a solution
    I followed the first set of equations, finding the net force on the spring in motion and then finding acceleration to get angular and then linear velocity.
    Then I followed the second set of equations using conservation of energy (assuming all of the PE at rest is converted to KE in motion).
    I got two similar, but different answers. Which way is the right way? Have I done anything right at all??
     
  2. jcsd
  3. Aug 16, 2012 #2

    cepheid

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    The problem with the force is that it's not constant, it's a function of x. That'll make it not so easy to solve for the position vs. time.

    Use energy methods, but don't forget gravitational potential energy!
     
  4. Aug 16, 2012 #3
    So initially my potential energy would come from the potential energy stored in the spring and the potential energy of gravity. But then I can assume that all potential energy is converted into kinetic energy correct?
     
  5. Aug 16, 2012 #4

    cepheid

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    Uh, yeah, sort of. It really depends on where you decide that each of the potential energy terms is equal to 0. It doesn't matter where: you get to choose this. Only changes in potential energy are significant, not specific values at specific points. So what you really have to do is figure out by how much the potential energy has changed in moving from the first height down to the second height, which is 0.4 m below. Since energy is conserved, any change in potential energy also occurs with an opposite change in kinetic energy.
     
  6. Aug 16, 2012 #5
    So in essence, my simplified equation would look like this.

    PE (elastic) + PE (gravity, initial) = KE + PE (gravity, final)

    I know that is way oversimplified but does my conceptual understanding see ok?
     
  7. Aug 16, 2012 #6

    cepheid

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    Most people define the elastic PE to be zero when the spring is unstretched and uncompressed (which is true at the beginning in our case), and then at the end of the drop, since the spring is now stretched, it has gained some elastic PE.
     
  8. Aug 16, 2012 #7
    Ohhhhhhh. I see what you mean.

    There is one other point on which I am still confused however. So since there is no elastic PE initially, then there should only be gravitational PE. But again, there is no displacement so that should also be zero? That is where I am stuck now.

    Would it be accurate to say this: ΔPE = ΔKE

    I think that is what you are getting at in saying that we are only concerned with the change in the gravitational energy.

    Thanks so much for you patience! I appreciate the help, physics doesn't come easy.
     
  9. Aug 16, 2012 #8

    cepheid

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    What do you mean by "there is no displacement?"

    Yeah, that's conservation of mechanical energy. EDIT: Actually you need ΔPE = -ΔKE

    What I'm getting at is that we are only concerned with the change in potential energy, not in the particular values it has before or after (which are irrelevant since you can add an arbitrary constant to them without changing the result). That having been said, gravitational potential energy is not the only one that changes. The elastic potential energy changes during the drop too.
     
  10. Aug 16, 2012 #9
    First, I was silly in saying "there is no displacement?" There obviously is displacement. That's a blonde moment for me.

    I understand that there will both be a change in the gravitational PE and the elastic PE. Here is a rough sketch of my equations. I am afraid I may be digging myself deeper into this mess, but I digress.

    ΔPE = ΔKE

    PE(final) - PE(initial) = -[KE(final) - KE(initial)]

    KE(initial) is zero.
    PE(final) is made of both gravitational and elastic PE.
    PE(initial) is made up of only gravitational PE.
    Thus...

    (1/2*k*x^2 + m*g*x(final)) - (m*g*x(initial)) = 1/2*m*v(final)^2

    If this is correct, then my only remaining confusion is the initial x position from which I am to obtain the initial gravitational PE?
     
    Last edited: Aug 16, 2012
  11. Aug 16, 2012 #10
    Sorry if I seem to be utterly confused.
     
  12. Aug 16, 2012 #11

    cepheid

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    That looks perfect. Regarding your question: again (and this is the point I was trying to make before), it doesn't matter what xinitial or xfinal are. All the matters is the difference between them. You'll notice that that's all that appears in your equation above. This difference will always be 0.4 m, because that's how much the height changed. If you decide that x (and therefore gravitational potential energy) should be 0 at the centre of the Earth (i.e this is the reference point for altitude measurements), Then you might have something like xinitial = 6400.0104 km, and xfinal = 6400.0100 km. (These are just example values: they could be anything as long as they differ by 0.4 m). The difference between them would still be 0.0004 km = 0.4 m. On the other hand, if you're more sensible and decide that x = 0 at the surface of the Earth, then you'll have xinitial = 10.4 m and xfinal = 10.0 m. The difference between them is still 0.4 m. You're talking about the same two points in space, just assigned coordinates relative to two different origins in these two examples.

    I guess another thing to point out is that you decided that the downward direction is positive (which is perfectly valid), which means that your x-values increase downward (unlike mine above), which means that your change in x is +0.4 m, unlike mine above, which was -0.4 m. Again, all of this is perfectly fine. You just have to pick a direction for your coordinate axis and stick to it.
     
  13. Aug 16, 2012 #12
    Aha! I get it now. Sorry to make you repeat it so many times. I can do this on my own now :) Thanks so much!

    I will be sure to re-check my positive and negative directions. :)
     
  14. Aug 16, 2012 #13

    cepheid

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    You also forgot to carry your negative sign through to the next line. It's important. if PE increases then KE decreases, and vice versa.
     
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