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Oscillation of a vertical spring

  1. Mar 3, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    A mass m hangs in equilibrium at the lower end of a vertical spring of natural length a, extending the spring to be a length b.

    1) Show that the frequency for small oscillations about the point of equilibrium is ##\omega = \sqrt{g/(b-a)}##

    2) The top end of the spring is made to oscillate vertically with a displacement ##c\sin (nt)##, moving downwards at t=0. Show that as a result, the length of the spring L(t) obeys: $$\ddot{L} + \omega^2 L = \omega^2 b + cn^2 \sin (nt)$$

    3. The attempt at a solution

    1)Write the eqn of motion: ##m\ddot{y} = -mg + k(y - a)##. In equilibrium, y = b, so ##mg = k(b-a) \Rightarrow k = mg/(b-a)##. Taking the period of a mass on a spring in the case of small oscillations, ##T = 2 \pi \sqrt{m/k}##, I get that ##T = 2 \pi \sqrt{(b-a)/g} \Rightarrow \omega = 2 \pi/T = 2 \pi \cdot \sqrt{g/(b-a)} \frac{1}{2 \pi}## which gives the result. Is this ok - am I allowed to simply quote the formula for the period of a mass on a spring? (I assumed yes, since it is given in the question that we are interested in small oscillations, so the formula is valid).

    2) If it is made to oscillate vertically with that displacement then the spring must be under some sort of external driving force. I haven't been able to make much progress with this one. I know the displacement is given by y = y(t) = c sin (nt). This means the external driving force causing this is of the form ##mcn^2 sin(nt)##. So ##m\ddot{L} = mg - k(L - c \sin(nt)) - mcn^2 \sin(nt)##, but this does not resemble what I need to show.

    Many thanks.
     
  2. jcsd
  3. Mar 3, 2013 #2
    Do not assume any force on the top end. Denote its location with some variable, say z. Denote the location of the other end with x. What forces act on it? What is its acceleration?
     
  4. Mar 3, 2013 #3

    CAF123

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    Is the spring attached to a wall at the top end? What is the 'it' you are referring to? The only forces that act on the spring is the external force (that makes it undergo that displacement), and a restoring force. (not gravity on the spring since it is not assigned a mass).
     
  5. Mar 3, 2013 #4
    The top end is attached to some mechanism that makes it oscillate as indicated. Denote the location of the top end as z. Do not analyze what force is required to make it move that way, just assume it does move that way. The other end of the spring has point mass m attached to it. Denote its location as x. Analyze the forces acting on the mass.
     
  6. Mar 3, 2013 #5

    CAF123

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    Ok, so the mechanism causes the mass to be displaced by an amount ##c \sin (nt)##. We are given that the natural length of the spring is a and it has some length L. So then writing the forces on the mass: ## m\ddot{L} = k(L - c\sin (nt)) -mg##
     
  7. Mar 3, 2013 #6
    That is close, but not quite. Specifically, why do attach the mass to the second derivative of the length? It is the mass whose acceleration your trying to find here. I personally find it easier to get the basic equation in terms of the two ends' positions first, then convert to L(t).
     
  8. Mar 3, 2013 #7

    CAF123

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    Actually, I think what I have written makes no sense at all since the mass of the spring is negligible. If I denote the position of the mass as x, then ##m\ddot{x} = k(L - c\sin(nt)) -mg##. I don't think we are interested in x, it is L that we want. What would be my next step?
     
  9. Mar 3, 2013 #8
    This term is wrong. What you have in brackets must be extension from the spring's natural length a.

    Once you get the rhs right, the next step would be to express the lhs in terms of L and the top end's motion.
     
  10. Mar 3, 2013 #9

    CAF123

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    Ok, so I should have ##\ddot{x} = \frac{k}{m}(a - c \sin(nt)) - g##

    Is the acceleration of the mass only slightly greater than the acceleration of the spring? So ##\ddot{L} = \ddot{x} - g##? For the top end motion, I am not really sure how to analyze here. The acceleration (I think) is ##cn^2 \sin(nt)##
     
  11. Mar 3, 2013 #10
    How come this is independent of the bottom end's motion?

    z is the top end's position. Then x(t) = z(t) - L(t).
     
  12. Mar 3, 2013 #11

    CAF123

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    In equilibrium, x = b without that driving force, so I think my expression should become: $$\ddot{x} = \frac{k}{m} (x - a - c\sin (nt)) - g$$


    I know that x represents the position of the mass, L(t) the length of the spring but is z simply the position of the end of the spring?

    So then, $$\ddot{z} - \ddot{L} = \frac{k}{m}(x - a - c\sin(nt)) - g, $$ where ##\ddot{z} = -cn^2 \sin(nt)##
     
  13. Mar 3, 2013 #12
    Note the description says "moving downward at time t = 0". So is z = c sin nt or z = - c sin nt?
     
  14. Mar 3, 2013 #13

    CAF123

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    z would be -c sin(nt). So ##\ddot{z} = cn^2 \sin(nt) ##.

    My eqn is then:$$\ddot{z} - \ddot{L} = \frac{k}{m}( - L - a ) - g$$and then I should just reexpress in terms of ##\omega##?
     
    Last edited: Mar 3, 2013
  15. Mar 3, 2013 #14
    Again, the term in the brackets must be the extension from length a.
     
  16. Mar 3, 2013 #15

    CAF123

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    I have the eqn in the form: $$\ddot{L} - \omega^2L = \omega^2 b + cn^2 \sin (nt),$$ but I appear to have a minus error in the second term on the LHS.

    Using the form given in the question, I proceeded to find L. The general soln to the homogeneous eqn is ##L_H(t) = A\cos(\omega t) + B \sin (\omega t). ## I am thinking to find the particular solution that I should find the soln to the two eqns ##\ddot{L} + \omega^2 L = \omega^2 b## and ##\ddot{L} + \omega^2 L = cn^2 \sin(nt)## separately and then make a linear combination of the two solns.

    However, in order to find the two parameters, A and B, I require two initial conditions. I think one is L(0) = -x and the other ##\dot{L}(0) = -nc?##
    Thanks.
     
  17. Mar 3, 2013 #16

    CAF123

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    All I did was replace x by z - l, z = c sin(nt) into the eqn at the bottom of post 11. Why is my eqn in post 11 wrong?
     
  18. Mar 3, 2013 #17
    Re the sign error, see #14.

    I am not sure what initial conditions you are supposed to have. If the assumption is that the point mass is in equilibrium at t = 0, then you conditions seem correct.
     
  19. Mar 3, 2013 #18
    Assume there is no sine term in the equation. Then it should be the usual harmonic oscillator equation. Is it the case?
     
  20. Mar 3, 2013 #19

    CAF123

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    If there is no sin term, then I have -k(x-a) on the RHS. Since x>a, then this represents an extension, so it is in the form of Hookes Law, no?
     
  21. Mar 3, 2013 #20
    x > a is incorrect, because x < 0 and a > 0.
     
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