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Math Help - Vertical Oscillations

  1. #1
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    Vertical Oscillations

    How do I tackle this question:

    A particle P of mass 0.3kg is hanging in equilibrium attached to one end of a light elastic string of natural length (l) of 0.6m and modulus (  \<br />
\lambda <br />
\) 9N. P is pulled downwards a further 0.4m and released. Calculate the time that elaspses before the string becomes slack, and the further time before the string is taut once more.

    For the first part of this question, I tried to solve it as follows:
    At equilibrium:
    \<br />
T = mg = \frac{{\lambda e}}{l}<br />
\
    where T is the tension in the string; and e is the extension at equilibrium.

    I found that e=0.196m
    And when P was pulled down a further 0.4m
    Using F=ma I obtained
    \<br />
mg - \frac{{\lambda (x + e)}}{l} =  - m\ddot x<br />
\
    after substituting in the values I got:
     \<br />
\ddot x =  - 50x<br />
\ where x is the extension of the string.
    Then I tried to use the equation
    \<br />
x = A\sin (\omega t)<br />
\ where A is the amplitude, t is time.
    When the string becomes slack, the extension is equal to the amplitude A, so I obtain the equation
     \<br />
0.4 = 0.4\sin (t\sqrt {50} )<br />
\
    giving me t=0.222s, but the answer in my book is 0.295s.

    Where have I gone wrong? Any help would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by free_to_fly View Post
    How do I tackle this question:

    A particle P of mass 0.3kg is hanging in equilibrium attached to one end of a light elastic string of natural length (l) of 0.6m and modulus (  \<br />
\lambda <br />
\) 9N. P is pulled downwards a further 0.4m and released. Calculate the time that elaspses before the string becomes slack, and the further time before the string is taut once more.

    For the first part of this question, I tried to solve it as follows:
    At equilibrium:
    \<br />
T = mg = \frac{{\lambda e}}{l}<br />
\
    where T is the tension in the string; and e is the extension at equilibrium.

    I found that e=0.196m
    And when P was pulled down a further 0.4m
    Using F=ma I obtained
    \<br />
mg - \frac{{\lambda (x + e)}}{l} =  - m\ddot x<br />
\
    after substituting in the values I got:
     \<br />
\ddot x =  - 50x<br />
\ where x is the extension of the string.

    Mr F says: You've forgotten the weight force down, chum. The equation of motion is

     \<br />
\ddot x =  9.8 - 50x , subject to the initial condition t = 0, \dot x = 0 and x = 0.196 + 0.4 = 0.596 m (see main reply).

    Note: x is the extension of the string measured from the unstretched position (position of zero extension), not the equilibrium position.


    Then I tried to use the equation
    \<br />
x = A\sin (\omega t)<br />
\ where A is the amplitude, t is time.
    When the string becomes slack, the extension is equal to the amplitude A, so I obtain the equation
     \<br />
0.4 = 0.4\sin (t\sqrt {50} )<br />
\
    giving me t=0.222s, but the answer in my book is 0.295s.

    Where have I gone wrong? Any help would be greatly appreciated.
    At equilibrium, (9.8)(0.3) = 15x => x = 0.196 m.

    When (after a bit of work) you solve the correct equation of motion, you have:

    x = 0.4 \sin \left( \sqrt{\frac{9.8}{0.196}} t + \frac{\pi}{2} \right) + 0.196 = 0.4 \sin \left(7.071 t + \frac{\pi}{2} \right) + 0.196.

    String is slack when x = 0:

    0 =  0.4 \sin \left(7.071 t + \frac{\pi}{2} \right) + 0.196

    \Rightarrow t = \frac{3.65368 - \frac{\pi}{2}}{7.071} = \, book's answer.

    Note: \sin^{-1}(-0.49) = \sin^{-1}(0.49) + \pi = 3.65368 is the value you want.
    Last edited by mr fantastic; February 13th 2008 at 12:54 AM.
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