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Need some help with spring-mass-damper problem

  1. Nov 1, 2004 #1
    Hi.

    I'm working with a spring-mass-damper-system problem but the thing is.. i can't figure out whether it is a single degree of freedom system(SDOF) or multi-degree of freedom system(MDOF). The question begins with a seat that is supported by a spring and also a damper. A man of 60 kg then sits on it. If the man slips off the seat withour imparting any vertical impulse to it, determine the expression for the ensuing motion of the seat against time.

    Is the 60kg weight counted as an excitation force? Any form of help will be greatly appreciated.

    Regards,
    Joel
     
  2. jcsd
  3. Nov 1, 2004 #2

    Tide

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    This is a one dimensional problem. Basically, they want to you determine what happens if the mass on an oscillating spring system changes abruptly.
     
  4. Nov 1, 2004 #3
    Okay..i'm still lost. Do i need to find out the natural frequency of the system?
     
  5. Nov 1, 2004 #4

    Tide

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    Of course! The presence of the man tells you how much the spring is compressed initially and now you have an oscillator starting with that amount of compression with only the seat itself as the mass. The system will undergo damped harmonic oscillation and ultimately approach a new equilibrium.
     
  6. Nov 1, 2004 #5
    Hmm but do i calculate the natural frequency using the spring stiffness or the damping coefficient, C?
    In the case of spring stiffness, natural freq, w = squareroot(K/m)
    In the case of damping coefficient, natural freq, w = C/2*m*damping factor,
    I assume its not the latter since i'm not given the damping factor.
    Or do we calulate the natural frequencies of the combined system of the man and the seat together?
     
  7. Nov 1, 2004 #6

    pervect

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    It all becomes clear if you write down the differntial equaiton of the system and solve it.

    You can write the force equation (with x = position, x'=velocity, x''=acceleration) as

    mx'' = -dx' - kx

    or mx'' + dx' + kx = 0

    the solution of which is [tex] x = C e^{\frac{-d +/-\sqrt{d^2-4 m k}}{2m} t}[/tex]

    This oscillates only if d^2-4 m k < 0, in which case the magnitude of the imaginary part is sqrt(d^2 - 4*m*k)/2m). You can see that in the limit as d=0, the oscillation frequency is sqrt(4*m*k)/2m = sqrt(m/k).
     
  8. Nov 1, 2004 #7
    I was given Damping coeeficient = 355 ,mass of seat = 40kg, mass of man = 60kg and spring constant = 19620.
    Would my equation of motion be 40x'' = -355x' - 19620x + 60*9.81 (inclusive of the man's weight)? This problem would be relatively easy but the addition of the man sitting down and slipping off the seat part makes it all difficult to understand for me. But i do realise this is a SDOF problem.
     
  9. Nov 1, 2004 #8

    pervect

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    re-write your equation as

    mx'' + dx' + k(x-a) = 0

    then substitute u=x-a

    You'll get rid of the constant term. This change of variables represents the measurement of the distance x as a departure from it's equilbrium position.
     
  10. Nov 2, 2004 #9
    Umm, i understand x is the vertical displacement of the seat from its equilibrium position, but wot is a?
     
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