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Energy in a spring-mass system

  1. Nov 2, 2008 #1
    1. The problem statement, all variables and given/known data
    A horizontal spring-mass system has low friction, spring stiffness 250 N/m, and mass 0.4 kg. The system is released with an initial compression of the spring of 13 cm and an initial speed of the mass of 3 m/s.
    (a) What is the maximum stretch during the motion?

    (b) What is the maximum speed during the motion?

    (c) Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system. What is the average power input in watts required to maintain a steady oscillation?
    watt



    3. The attempt at a solution

    i was able to solve a and b right but having problems with c. Thanks for looking.
     
  2. jcsd
  3. Nov 2, 2008 #2
    Well, figure out how long it takes a cycle to happen, (the period) and then divide by 0.02 and this will be the joules dissipated per second.

    Then this will be the amount of joules you need to keep pumping into the system to keep it going. (Remember watts is joules per second).

    Does that sound reasonable?
     
  4. Nov 2, 2008 #3
    Thanks for your reply. So do i find W which is equal to sqrt(K/m) and from there find the Period and do the rest? Thanks
     
  5. Nov 2, 2008 #4
    Yeah. And remember:

    [tex] \omega = 2\pi f [/tex]

    [tex] T = \frac{1}{f} [/tex]

    That should be enough to get to the answer.
     
  6. Nov 2, 2008 #5
    Sorry I said before to divide the period by 0.2, I meant divide 0.2 by the period. If you look at it in terms of units:

    [tex] \frac{\frac{J}{cycle}}{\frac{s}{cycle}} = \frac{J}{cycle} X \frac{cycle}{s} = \frac{J}{s} [/tex]

    Remembering when dividing fractions you just multiply by the inverse of the bottom fraction...
     
  7. Nov 2, 2008 #6
    I calculated W = sqrt(250/0.4) and i got 25 for w. Then w = 2*pi*f and got f = w/2*pi = 3.978 then 1/f to get T so T = 0.251 and i divided that by 0.02 and got 12.56 but the answer was incorrect.
     
  8. Nov 2, 2008 #7
    Thank you got it
     
  9. Nov 2, 2008 #8
    Sorry I misled you... see previous post.

    My answer was 0.8 watts.
     
  10. Nov 2, 2008 #9
    Thanks man your awsome.
     
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