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Spring from a ceiling problem

  1. Jan 15, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    A massless spring with no mass attached to it hangs from the ceiling. Its length is 0.2
    meter. A mass M is now hung on the lower end of the spring. Imagine supporting the mass
    with your hand so that the spring remains relaxed (that is, not expanded or compressed),
    then suddenly remove your supporting hand. The mass and spring oscillate. The lowest
    position of the mass during the oscillations is 0.1 m below the place it was resting when
    you supported it.
    (a) What is the frequency of oscillation?
    (b) What is the velocity when the mass is 0.05 m below its original resting place?

    2. Relevant equations
    t = sqrt(2*d/g)
    g = 9.8
    frequency = w/2pi
    w^2=g/l
    x=Acos(wt+Φ)

    3. The attempt at a solution

    These problems always throw me off. I do not want to be thrown off this time around! Please check my work. I am somewhat confident in it.

    1. To find frequency of oscillating we find w. w is easy enough by the equation w^2=g/l. Since the length is 0.2 and g = 9.8 this comes out to 7.
    question: I used the correct length correct? Can I use this equation? This was giving to me in my pendulum notes but we are working with a spring... Is this incorrect?

    2. With w, it is easy to see frequency = w/2pi, so 7/2pi is thus the frequency and part a.

    3. To find velocity we need the position function which is x = 0.1sin(7*t)
    or equally 0.1cos(7t+pi/2)

    4. Take the derivative which becomes velocity = 0.7cos(7t)

    5. Find t for 0.05m. t=sqrt(2*0.05/9.8) = 0.101

    6. Plug t into velocity equation to get 0.7cos(7*0.101) = 0.6999 = 0.7m/s

    Did I use wrong equations and get the wrong answer? I am somewhat skeptical about my final velocity as it is saying that cos(0)=1 and so it takes on 0.7.
     
  2. jcsd
  3. Jan 15, 2016 #2
    You need to first use the given info and compute the spring constant. Then use the spring constant and the mass to solve the rest of the problem.

    Don't skip to the solution to the equation of motion. Use Newton's second law to express the equation of motion.

    This will allow computing the frequency in terms of mass and spring constant.

    A picture never hurts either.
     
  4. Jan 15, 2016 #3

    RJLiberator

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    We do not have the mass, otherwise I would have done that for sure. Is there something I am missing here?
     
  5. Jan 15, 2016 #4

    haruspex

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    A bit of dimensional analysis helps here. You know g, and you are given some distances, so all of the information you have involves distances and times only. That may well be enough to find a frequency and a velocity, since those also involve only distance and time. Knowing the mass cannot help unless you also know something else that involves a mass dimension, such as the spring constant. So there is a very good chance that you do not need to know either.
    Proceed by letting the mass be M and deriving equations involving M. Expect M to disappear later.
     
  6. Jan 15, 2016 #5

    TSny

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    The system is not oscillating as a pendulum. So you are using the wrong equation for the frequency.

    You will not be able to determine the mass or the spring constant. However, you should be able to determine the ratio k/m, which is all you need to answer the questions.
     
  7. Jan 16, 2016 #6
    This plan nearly always succeeds in intro physics when you do not have the mass and think you need it.
     
  8. Jan 20, 2016 #7

    RJLiberator

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    Boom. Been starring at this problem for the past few days, I think I finally have it.

    Hooke's Law: F = -kx
    k = F/x when looking at just magnitudes
    since F = m*g
    k = mg/x

    now, k/m = g/x = w^2
    g = 9.81 m/s^2
    x = 0.2m (Am I using the right measure for x here? 0.2m length is the resting length of the string before mass was added, from what I see, that is the number we are supposed to use. Or am I to use 0.2m + 0.1m = 0.3m as the stretched out length??)

    so w^2 = 9.81/0.2
    w = sqrt(9.81/0.2) = 7.00 s^-1
    this agrees with my first calculation of w.

    Did I do something wrong here or is this the right approach?
     
  9. Jan 20, 2016 #8

    TSny

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    I'm not sure of what you did here. Why did you say that the gravitational force is equal to the spring force? Is this always true or is it true for only one particular value of x?

    To make sure that you have the correct interpretation of x in the formula F = -kx, can you state the value of x at the initial position where the block was released?

    What is the value of x when the block gets to its lowest point?
     
  10. Jan 20, 2016 #9

    RJLiberator

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    We have some mass, M hanging from a massless spring from the ceiling. The massless spring is initially 0.2 m long prior to attaching the mass. When the mass is attached and let to reach it's amplitude, it falls another 0.1 m down from the ceiling. This is a total of 0.3 m from the ceiling.

    I am using Hooke's law where k = F/x. Force in this problem is equal to m*a = m*g since gravity is the acceleration / only force acting on the mass here.
    Now, using the hints in this thread, I saw that k/m = g/x = w^2.

    The value of x at the initial position was 0.2 m long. The resting point for the spring is 0.2 m. That's why I think that value is the x value.

    If we let the equilibrium position be x = 0, then the amplitude is 0.1m and the lowest point is x = 0.1m or -0.1m depending on the orientation of your plot.

    :) ?
     
  11. Jan 20, 2016 #10

    haruspex

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    You need to be clearer in your mind (and preferably in your postings too) regarding what your variables represent.
    The F in Hooke's law refers to the tension. The F in F=mg is the gravitational force. Under what circumstances will these be equal? What does that imply for the meaning of x in the last equation above?
     
  12. Jan 20, 2016 #11

    RJLiberator

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    I see.
    That is a subtle problem.

    So, in actuality, the x value should be the change in x. The change in x is 0.1m and NOT 0.2m which I had originally thought was the case.
     
  13. Jan 20, 2016 #12

    haruspex

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    Closer, but still not right. Please try to answer the question I posed in the previous post: under what circumstances will the two forces be equal?
     
  14. Jan 20, 2016 #13

    RJLiberator

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    I think I have it:
    It is when we are at equilibrium. Here the tension is equal to the force applied when it is in equilibrium
    And therefore, 0.2m was actually the correct x value as we are describing the system at equilibrium and using the equations giving by it.
     
  15. Jan 20, 2016 #14

    haruspex

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    Yes, the x you calculated is the equilibrium extension with the mass suspended. But that is not 0.2m. We do not know what it is yet.
    You have:
    Relaxed length = initial length before release = 0.2m.
    Maximum extension = 0.1m.
    Equilibrium extension = x.
    What is the relationship between the three extensions?
     
  16. Jan 20, 2016 #15

    RJLiberator

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    Oh man, classical mechanics is the death of me!
    Such a joy, but wow, how many times can I be wrong. ^_^.

    So what is happening is that when the mass is attached, it actually reaches the lowest point of 0.3m spring length.
    So this is the number I should be using.
    This makes sense, because that is where the system is at rest, 0.3m stretched spring where the forces cancel each other out.
     
  17. Jan 20, 2016 #16

    haruspex

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    No, 0.1m is the maximum extension after the system is released from the relaxed length.
    If the system oscillates without loss, what is the minimum extension in each cycle?
    Given the minimum and maximum extension each cycle, what is the equilibrium extension?
     
  18. Jan 20, 2016 #17

    RJLiberator

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    Well, 0.1 is the amplitude, so 0.1+0.1 = 0.2 as the oscillation cycle displaced. (total)

    So the minimum extension should also be 0.1 since the system goes on without loss, correct?

    The equilibrium extension is 0.1m away from equilibrium.

    How do we get a different number than 0.1m as amplitude or extension if nothing is lost and this is SHM?
     
  19. Jan 20, 2016 #18

    haruspex

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    what is the minimum extensions each cycle?
    What is the amximum extension each cycle?
    So what is the amplitude?
     
  20. Jan 20, 2016 #19

    RJLiberator

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    You pose a question that I am having trouble visualizing, let me explain.

    Maximum extension: 0.1m
    This is brought on by the fact that it reaches this distance as its low point per the question statement.
    Minimum extension: 0.1m or 0m.
    Here, perhaps I am not understanding the question. How is there a minimum extension each cycle? If there is no loss here, then the oscillation will be simple harmonic motion and continue on going for 0.1m below the initial position to 0.1m above the initial position, continuing. This would imply that hte minimum extension is = maximum extension = 0.1m. Perhaps you mean that the minimum is when it was at rest at 0.2m ?
    Amplitude: 0.1m

    If we look at a diagram, we see the start = 0.2m, the low point = 0.3m the return heigh point = 0.1 m. Is this what you mean?
     
  21. Jan 20, 2016 #20

    haruspex

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    How will it go above the release position? Think about energy.
     
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