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Mass hanging by spring tracing a eight shaped curve

  1. Apr 22, 2014 #1
    1. The problem statement, all variables and given/known data
    A mass ##m## on the end of a light spring of force constant ##k## stretches the spring to a length ##l## when at rest. The mass is now set into motion so it executes up and down vibrations while swinging back and forth as a pendulum. The mass moves in a figure-eight pattern in a vertical plane, as shown in the figure. Find the force constant in terms of ##m##,##l## and ##g##.

    (Ans: k=4mg/l )

    2. Relevant equations



    3. The attempt at a solution
    I noticed that the curve traced by the hanging mass is of the form ##r^2=a\cos(2\theta)## with the mass ##m## being at the origin at ##t=0##. But I don't think this is going to help. This is a question from one of my practice sheets and I doubt I need to deal with polar curves which aren't generally taught in high school.

    How do I approach this problem? :confused:

    Any help is appreciated. Thanks!
     

    Attached Files:

    Last edited: Apr 22, 2014
  2. jcsd
  3. Apr 22, 2014 #2

    jedishrfu

    Staff: Mentor

    Since it traces out a figure eight that would suggest that one oscillation is twice what the other oscillation is so maybe you can factor that into your solution.
     
  4. Apr 22, 2014 #3

    andrevdh

    User Avatar
    Homework Helper

    Lets me think of Lissajous figures. For a figure 8 the frequencies of oscillations are 1:2. In this case the up-down oscillations are twice those of the pendulum motion. Not sure this helps or is relevant though.
     
  5. Apr 22, 2014 #4
    Ah yes, thanks a lot jedishrfu! :)

    $$2\times 2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{l}{g}}$$
    $$\Rightarrow k=\frac{4mg}{l}$$
     
  6. Apr 22, 2014 #5

    andrevdh

    User Avatar
    Homework Helper

    I think the springs' frequency of oscillations is twice that of the pendulum in this case.
     
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