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Equation of motion: spring mass system - free undamped vibration

  1. Sep 7, 2012 #1
    1. The problem statement, all variables and given/known data

    i have uploaded my question. please check out the attached .pdf file.


    2. Relevant equations



    3. The attempt at a solution
     

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  3. Sep 7, 2012 #2

    Doc Al

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    Shouldn't the 'inertial force' oppose the acceleration, not the motion?
     
  4. Sep 7, 2012 #3
    hey! thanks for replying.

    i was following this website. here's the link

    http://lpsa.swarthmore.edu/Systems/MechTranslating/TransMechSysModel.html

    it has been mentioned in this website that 'inertia force acts opposite to the direction of motion'.

    did i understand it right?
    mass will decelerate in the positive x direction, when it moves away from equilibrium towards +A.
    it will accelerate in the negative x direction, when it is moving towards the equilibrium position.
    it will start decelerating in the negative x direction, when it moves away from equilibrium position towards -A.
    it will start accelerating in the positive x direction, when it moves towards equilibrium position.
     
    Last edited: Sep 7, 2012
  5. Sep 7, 2012 #4

    Doc Al

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    I think it's just a sloppy and misleading way to describe the direction. Note how they 'derive' the inertia force: It's just the negative of 'ma'.

    Since your variable x already defines the positive direction of acceleration, [itex]\ddot{x}[/itex], the inertia force is simply [itex]-m\ddot{x}[/itex]. The direction is already included: When the sign of [itex]\ddot{x}[/itex] changes, so will the sign of the inertia force.

    But it's the direction of the acceleration that determines the direction of the inertia force, not the direction of motion.
     
  6. Sep 7, 2012 #5
    as i defined downward x as positive, so does that mean x¨ is positive as i move downwards away from equilibrium?

    i might be asking really very silly doubt but this thing is really confusing me. i want it to be sorted.

    as the mass moves away from equilibrium towards +A in the downward positive x direction, if we draw a free body diagram, how do we represent acceleration?

    if we take x¨, acceleration positive in the downward direction, the inertia force acts in the opposite direction i.e, upwards.

    so if i write equation of motion using D'Alembert's Principle, i get:

    -mx¨ - kx = 0.

    but if i consider the situation where the mass is at a position, away from -A, towards equilibrium, then what is the direction of acceleration and how do i write the equation of motion?
     
    Last edited by a moderator: Sep 7, 2012
  7. Sep 7, 2012 #6

    Doc Al

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    [itex]\ddot{x}[/itex] is positive as long as the acceleration is in the positive direction, which you have defined as downward. Note that the restoring force of the spring always acts to accelerate the mass back towards the equilibrium point. So at any point below the equilibrium point, the acceleration would be upward.

    You wouldn't put the acceleration on a free body diagram, but it would act upward.

    [itex]\ddot{x}[/itex] would be negative and the inertia force would be positive. But you don't have to worry about the inertia force direction--you'd still represent that inertia force as [itex]-m\ddot{x}[/itex].

    Right.

    So now you are looking at the situation when the mass is above the equilibrium point. So the spring force acts downward and the acceleration is downward.

    Nonetheless, you'd still write the restoring force as -kx (since x is now negative, the force comes out positive). And you'd still write the inertial force as opposite to the acceleration: [itex]-m\ddot{x}[/itex]. (Since the acceleration will be positive, that will come out to be negative.)

    So the equation of motion will be exactly the same. Realize that you don't have to know the direction of the acceleration ahead of time. Just represent the inertia force properly--with respect to your chosen coordinates. Then your equation of motion will tell you the direction of the acceleration.
     
  8. Sep 8, 2012 #7
    thanks Doc Al. i think i got it.

    if i have any doubts further, i'll get back to you.
    thank you very much!
     
  9. Sep 8, 2012 #8

    Doc Al

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    Excellent. Glad that it helped. And you are most welcome.
     
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