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Understanding spring mass system

  1. Apr 19, 2005 #1
    I am just learining basic dynamics from spring mass system.
    there is a horizontal spring fixed to the wall and a mass is at the end.
    When the mass is displaced by a force F to the right the spring stretches to a distance X. Now i draw the STATIC FBD. The
    eqn is f-kx=net force on the mass. Now i draw the kinetic diagram and the equation i derive is

    f-kx=ma where a is the acceleration of the mass m. So for this spring mass system the dynamic equation is

    Is what i have done is true and correct?
  2. jcsd
  3. Apr 19, 2005 #2


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    Is the force acting continuously on the object...?Or it just stretches the string & lets it go to oscillate...?

    That equation,vector or scalar,f-kx-ma=0 makes no sense...


  4. Apr 19, 2005 #3
    Yes, it is correct as long the hand pulling the mass is not releasing it.

    [tex]F - kx = 0[/tex]
  5. Apr 19, 2005 #4

    Doc Al

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    This is just a statement that the net force on the mass equals f - kx.
    Now you've applied Newton's 2nd law: Net Force = ma. Note that rewriting it as "Net Force - ma = 0" adds nothing.

    If the mass is in equilibrium, then f - kx = 0; if not, then it will accelerate.
  6. Apr 23, 2005 #5
    Here, you can learn more on basic dynamics. make sure that you are able to solve the given exercises...


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