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Thread: Motion caused by elastic strings

  1. #1
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    Question Motion caused by elastic strings

    I'm working through an old UK A-Level Applied Mathematics textbook and am encountering a few problems. Here is one I'm struggling with - the second part of the question only.

    Two identical elastic strings of length 1 metre and modulus 4.9 N are each fastened to a particle of mass 0.5 kg. Their other ends are fixed to two points 4 metres apart in a vertical line. Find the height of the particle above the lower fixed point A in the equilibrium position. The particle is now pulled down to A and released from rest. Find the greatest height above A to which the particle rises.

    For the first part, here is my answer:

    Let the tensions in the strings be T1, T2 and let the extension in the upper string be x.
    The particle is in equilibrium, so T1 - T2 = 0.5g = 4.9
    Also, by Hooke's Law, T1 = 4.9x and T2 = 4.9(2 - x)

    Therefore 4.9x - 4.9(2 - x) = 4.9 so x = 1.5 m.

    So, the height of the particle above A in equilibrium is 3 - x = 1.5 m.


    Here is my attempt at the second part, using Conservation of Mechanical Energy. Although I get the same numerical answer as the book, they have it as 2√2 m, and I don't understand how they get that, so maybe I'm doing something wrong.

    Let h be the greatest height above A reached by the particle. Then considering the mechanical energy of the system:

    At the initial position: Elastic Potential Energy = 4.9 x 3^2/2 = 22.05 J

    When the particle reaches its highest point: Potential Energy = 0.5gh = 4.9h (taking g = 9.8 m s-2)
    Elastic Potential Energy Motion caused by elastic strings-eqn-1.jpg
    (because the lower string is pulling the particle down, whilst the upper string is pulling it up)

    Total mechanical energy is conserved, so 4.9h + 9.8(h - 2) = 22.05
    i.e. h = 17/6 = 2.83 m

    Now, to 2 dp this is the same numerical value as the solution of 2√2 m given in the book, but I'd really like to know how the authors got to that solution. What am I doing wrong?
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  2. #2
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    Re: Motion caused by elastic strings

    Now, to 2 dp this is the same numerical value as the solution of 2√2 m given in the book, but I'd really like to know how the authors got to that solution. What am I doing wrong?
    I do not agree with the book's "exact" solution ... I get $h = \dfrac{3+\sqrt{7}}{2} \approx 2.823 \, m$

    $\dfrac{1}{2}k \cdot 3^2 = mgh + \dfrac{1}{2}k(h-1)^2 + \dfrac{1}{2}k(3-h)^2$

    $k \cdot 3^2 = 2mgh + k(h-1)^2 + k(3-h)^2$

    since $4.9 = k = mg$ ...

    $3^2 = 2h + (h-1)^2 + (3-h)^2$

    expanding and combining terms yields the quadratic ...

    $2h^2 - 6h + 1 = 0$

    quadratic formula yields the result for $h$ given above,
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    Re: Motion caused by elastic strings

    Yes, agreed. Glad it wasn't just me! It's not the first error in the answers that I've found in that book. Thank you, Skeeter.
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