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Finding frequency of spring

  1. Apr 22, 2015 #1
    1. The problem statement, all variables and given/known data
    If I attach a spring to the ceiling, then hang a weight on the spring, the spring extends by 1.19 meters, comes to rest for an instant, and then begins oscillating. The mass of the weight is 6.6 kg. What is the oscillation frequency of the block's oscillations.

    2. Relevant equations

    upload_2015-4-22_20-10-25.png
    3. The attempt at a solution

    solving for k,

    F =kx
    mg = kx
    9.8m/s^2 x 6.6kg = kx
    64.68 kgm/s^2 / 1.19m = 54.35 = k

    plugging into the above equation

    T comes out to be 2.18

    and F = 1/T = 1/2.18 = .456 Hz

    is this correct?
     
  2. jcsd
  3. Apr 22, 2015 #2

    collinsmark

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    Actually, in this case, no.

    You don't want to use Hooke's law here*. When the weight block falls a total distance of 1.19 m, its velocity has momentarily dropped to zero, but its acceleration is at maximum. In other words, the net force on the block is certainly not zero; the force on the mass block from the spring is certainly greater than the weight's block's weight. Otherwise the weight block wouldn't shoot back up.

    In this case you can apply principles of conservation of energy to find the spring constant.

    *There is another trick to find the spring constant by using Hooke's law directly, if you can first figure out the distance at which the weight's block's acceleration, rather than velocity, is zero.** But conservation of energy might be a little more intuitive, and gives the same answer.

    [Edit: Try and see if you can calculate the spring constant using both methods. That way you can double check your answer! :wink:]

    **[Another edit: To help find out where this point is, ask yourself, "along this 1.19 m traveled, at what point does the weight block stop speeding up and start slowing down?" Symmetry might be useful here.]

    [A final edit: I went though my post and reworded my answer, crossing terms out where appropriate, calling the "weight" a "block." The problem statement calls it "a weight." But good golly that can get confusing when talking about the "weight's mass," "weight's weight," "acceleration of the weight," ... I mean good grief, what a horrible term to call the thing. So I just replaced the term with "block" when I'm talking about the physical thingy, and used the term "weight" only when I'm talking about how much the thingy weighs.]
     
    Last edited: Apr 23, 2015
  4. Apr 23, 2015 #3
    can we use mgh = 1/2 k x ^2?

    also, it has highest acceleration at the points where V is 0(at the lowest and highest points)

    hmm if we use mgh = 1/2 k x^2

    k comes out to be 4.82
     
    Last edited: Apr 23, 2015
  5. Apr 23, 2015 #4
    also, F=-kx can only be used when there is netforce = 0?
     
  6. Apr 23, 2015 #5
    I wouldn't recommend to do it this way. For one thing, Hooke's law assumes a static load when solving for k.
     
    Last edited: Apr 23, 2015
  7. Apr 23, 2015 #6
    No, you can use it anytime as long as the spring obeys Hooke's law.
     
  8. Apr 23, 2015 #7
    and when is hooke's look being obeyed or not? give some examples please
     
  9. Apr 23, 2015 #8
    It's usually assumed to be but it's a good idea to state this at the start.
     
  10. Apr 23, 2015 #9
    well the only 2 ways i know to solve for k is f = -ks and 1/2 k x ^2 = PE, so is k = 4.8 or 55
     
  11. Apr 23, 2015 #10
    I think the former equation is the way to go rather than the latter.
     
  12. Apr 23, 2015 #11

    SammyS

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    If done correctly, both of your methods work and will agree, as they should.

    The position of the end of the spring oscillates from some minimum height to some position 1.19 meters higher. At what location is the force exerted by the spring equal in magnitude to the force gravity exerts on that object? (I hate to call it a "weight" since that's also what we call the force that gravity exerts on the object.)
     
  13. Apr 23, 2015 #12
    I think i did that in the first attempt, mg = kx = F
    k = 54.35
     
  14. Apr 23, 2015 #13

    SammyS

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    You didn't really answer the question.

    Again:
    At what location is the force exerted by the spring equal in magnitude to the force gravity exerts on that object?​
     
  15. Apr 23, 2015 #14
    at the bottom, 1.19 meters from resting point , where it stops and acceleration is at its highest, pointing upwards.
     
  16. Apr 23, 2015 #15

    SammyS

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    It's not either of those locations.
    At the top, the spring exerts zero force.

    At the bottom, the object comes to rest. If the spring force upward was equal to the gravitational force downward, the net force on the object would be zero at the same time and place that it is stationary. It would then just stay there motionless.​

    Furthermore, you have previously stated (correctly) that the object's maximum acceleration is at the top & bottom. Newton II then implies the net force is maximum there.

    So, one more time:
    At what location is the force exerted by the spring equal in magnitude to the force gravity exerts on that object?​
     
  17. Apr 23, 2015 #16
    ok, i drew a drawing : upload_2015-4-23_19-32-56.png
    so at point C, isn't the spring exerting a force on the mass since it is decompressed?

    also to answer the question , the f net at the mid point (aka the point where the block is at rest) is 0 so that is where acceleration of the spring is 9.8 m/s^2 but in the opposite direction of gravity.
     

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  18. Apr 23, 2015 #17

    SammyS

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    No. The lower end of the spring is at it's equilibrium position. It is stretched zero from equilibrium, so it exerts no force.
    The mid point is indeed the place, but as far as the spring is concerned, acceleration of the spring means nothing. At that point, (the mid point), what is the acceleration of the object hanging on the end of the spring?
     
  19. Apr 23, 2015 #18
    in my drawing, point A is the spring at rest, then at B, i pull the spring downwards, then it goes past point A into C, then into A again, then back to B, then A, then C.

    so at point C, the only force acting on the block is gravity? doesn't a spring exert a force if it is compressed?
     
    Last edited: Apr 23, 2015
  20. Apr 23, 2015 #19

    SammyS

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    Ah Hah ! I think I see some of the problem you're having.

    As I read the problem statement: (Yes, there are some assumptions built into this.)

    A spring is attached to the ceiling. Of course, it hangs down relaxed, unstretched, but also uncompressed, with the free end at its equilibrium position.

    With the spring is in this state, an object is carefully attached to the end of the spring, fully supported some way. The object is then released and descends to a point which is 1.19 meters below the location at which it was attached to the spring, comes to rest for an instant, after which it begins oscillating .​

    The object does not go any higher than the point at which it was released. The spring is never compressed.

    At the mid-point the acceleration is zero so the net force on the object is zero. Furthermore, at this mid-point position, the object is not at rest, in fact its maximum speed is at mid-point.
     
  21. Apr 23, 2015 #20
    ok, i see, the object doesn't ever go past the point where it was dropped, but does it reach that point and then stop, then goes back to the bottom?
     
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