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Math Help - Linear motion problem

  1. #1
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    Linear motion problem

    When a mass is suspended from a string, the equilibrium length is increased by 50mm. the mass is then given a blow which starts it moving vertically at 200mms^1. Find the period and amplitude of the resulting oscillations, assuming no damping.
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  2. #2
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    Two questions:

    1. Did you mean "spring" instead of "string"?
    2. What have you done so far?
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  3. #3
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    sorry i meant spring. My problem is that i cant set up the model!
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    To obtain your model, you'll need to apply Newton's Second Law to the mass spring system. You know that F = ma. For a spring in regime of Hooke's Law, the force F = -kx. There is no damping or forcing, which simplifies matters greatly. The initial velocity is given. I think you can also assume that the spring starts in equilibrium. Those two facts give you the initial conditions for your DE. So, can you now state the DE that governs this system, including the initial conditions?
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    I get \ddot{x}+\frac{kx}{m}=g

    so x=x_0cos(wt)+\frac{v_0}{w}sin(wt) where the PI is \frac{mg}{k}.

    However if x_0=0,v_0=0 most of the DE goes to zero, this does not correlate to the answer in my textbook which is 0.477s and 14.2mm.
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  6. #6
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    I don't agree with your DE. Let's analyze the spring before the upward force is applied. It's in equilibrium, right? Let's say that after the mass m was applied, the new equilibrium position was \ell away from where it was before. Applying Newton's Second Law to the new equilibrium, the forces must negate each other. That is, the upward force of k\ell must equal the downward force of the mass mg. So, k\ell=mg. We'll tuck that away for later.

    Now, applying Newton's Second Law to the mass as it's moving. The spring force is -kx, where x is measured from the new equilibrium position. Gravity, believe it or not, is no longer in play. All gravity does is change the equilibrium position of the spring. You can actually prove that. Since we're measuring from the (gravity including) new equilibrium position, we need not include that term in the DE as a force when applying Newton's Second Law. It'll come back into the equation due to the k\ell=mg equation we found earlier.

    So, Newton's Second Law states that

    m\ddot{x}=-kx.

    But now k=mg/\ell, and hence the DE becomes

    \displaystyle{m\ddot{x}=-\frac{mg}{\ell}\,x,} or

    \displaystyle{\ell\ddot{x}=-gx.}

    Your initial conditions are not all zero, as you indicated. You have an initial upward velocity. That will save your solution from being the trivial solution.

    ...where the PI is \frac{mg}{k}...
    What is PI?
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  7. #7
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    sorry PI means particular integral but it is not needed if m\ddot{x}=-kx as the DE is a homogeneous one.

    I still struggle to see why it isnt mg-kx=m\ddot{x} as surely this is the resultant force acting on the mass.

    furthermore if m\ddot{x}=-kx then we have m\ddot{x}+kx=0 which goes to
    x=x_0cos(wt)+\frac{v_0}{w}sin(wt) where w=\sqrt\frac{k}{m}

    substituting in v_0=200,x_0=l
    we have x=lcos(wt)+\frac{200}{w}sin(wt)

    however i still cant find w
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  8. #8
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    Ok, here's why we can ignore gravity in the DE. If you solve the DE you think is correct (which, incidentally, is correct if you measure x from its pre-mass equilibrium position), then your DE becomes (after substituting in k=mg/\ell)

    mg-kx=m\ddot{x}

    \displaystyle{mg-\frac{mg}{\ell}\,x=m\ddot{x}}

    \displaystyle{g-\frac{g}{l}\,x=\ddot{x}}.

    The solution of this equation is

    \displaystyle{x=\ell+A\cos\left(\sqrt{\frac{g}{\el  l}}\,t\right)+B\sin\left(\sqrt{\frac{g}{\ell}}\,t\  right)}.

    You can see that, aside from a fixed displacement \ell, which we already knew about, the motion is SHM from the new equilibrium position. This means you could just as easily define a new coordinate y=x-\ell. That is,

    \displaystyle{y=A\cos\left(\sqrt{\frac{g}{\ell}}\,  t\right)+B\sin\left(\sqrt{\frac{g}{\ell}}\,t\right  )}.

    Now, just by differentiating twice, you can see that this coordinate y solves the following DE:

    \displaystyle{-\frac{g}{l}\,y=\ddot{y}},

    which is the DE I proposed (sans gravity).

    This new coordinate is measuring the displacement from the new equilibrium. Does this make sense now? It's not an a priori line of reasoning, I grant you. But I still think it's compelling. There does exist an a priori line of reasoning to show why you can just take the DE I proposed as accurately describing the system. However, I have forgotten what that argument is, and I don't have my mechanics book with me.

    In reply to the work you have done, I would say that you're not using all the information that you have. In particular, you're not using the value of k=mg/\ell. That's why you're not able to find all the values of the constants and such.

    Does all this make sense?
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  9. #9
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    I think ive just got mixed up with equilibrium position and i didnt see k=mg/l thanks
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  10. #10
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    You're welcome. Have a good one!
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