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Math Help - Damping ratio

  1. #1
    srh
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    Damping ratio

    Hi guys,
    Revision for an exam tomorrow, I hope someone can help?

    I've been given the answers but I can't work out how to find the damping ratio without being given a velocity or distance for the mass to travel?


    A mass of 6kg is suspended on a spring and set oscillating. it is observed that amplitude reduces to 5% of its initial value after 2 oscillations, which takes 0.57 seconds.

    Determine the following.

    The damping ratio (0.43)
    The natural frequency (3.51)
    The actual frequency (3.17)
    The spring stiffness (2915N/m)
    The critical damping coefficient (264.5Ns/m)
    The actual damping coefficient (113.7Ns/m)

    I can use the damping ratio to help solve some of the other values once I know how to work it out.

    Any help appreciated
    Last edited by srh; January 20th 2013 at 12:12 PM.
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  2. #2
    srh
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    Re: Damping ratio

    OK this is what I have,
    Natural frequency = 2/0.57 = 3.51Hz
    Spring stiffness 3.51Hz = 1/2pi x square root of K/m (k works out to be about 2915)
    Critical damping square root of 4mk = 264.5

    I need the damping ratio to find actual damping and actual frequency, I don't think I can do it without a velocity?
    The 5% of amplitude must come into it somehow?
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