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Homework Help: Spring Oscillation Problem

  1. Dec 9, 2011 #1
    1. The problem statement, all variables and given/known data
    A 200 gram block is attached to a spring with a spring constant of 8 N/m. The spring oscillates horizontally on a frictionless surface. Its velocity is 80 cm/s when x = - 4.2 cm.

    a. What is the amplitude of oscillation?

    b. What is the block’s maximum acceleration?

    c. What is the block’s position when the acceleration is maximum?

    d. What is the speed of the block when x = 2.5 cm?



    2. Relevant equations

    A= sqroot(Xo^2+(Vo^2/w^2))
    amax=w^2A

    3. The attempt at a solution

    I tried solving for Amplitude but Xo and Vo and w are not given, i tried looking for a way to find these variables but could not find one
     
  2. jcsd
  3. Dec 9, 2011 #2
    If it's frictionless then then total energy is constant. If you have the information of the system at one point, you can find the total energy; the maximum amplitude will occur when the kinetic energy is 0, and the energy inside the spring is at a maximum.
     
  4. Dec 9, 2011 #3

    gneill

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    Staff: Mentor

    Also, you should have a formula for the natural frequency ω of a mass spring system. The given displacement and velocity will work for your Xo and Vo.
     
  5. Dec 9, 2011 #4
    ω = 2pie(f) but i am not given the period or frequency
     
  6. Dec 9, 2011 #5
    unless frequency is dependant on 'k' and 'm'
     
  7. Dec 9, 2011 #6
    ω= sqroot(k/m) =6.32 therefore Amp=4.2 ?
     
  8. Dec 9, 2011 #7
    Looks wrong to me. root(8/0.2) seems like its going to be over 6.
     
  9. Dec 9, 2011 #8
    8/.2 = 40 root(40) is 6.32
     
  10. Dec 9, 2011 #9
    Oh I'm sorry! I only saw the 4.2. yes that looks right.
     
  11. Dec 9, 2011 #10
    i must convert all the conversions to meters instead of cm correct? which will give an Amplitude of .133?
     
  12. Dec 9, 2011 #11
    Yeah you'll have to convert all the units to be the same. It doesn't matter which way you go though. have you studied energy yet? I encourage (strongly) for you to double check your answers using the methodology that I explained in my first post.
     
  13. Dec 9, 2011 #12
    yes i have, i have the correct answers for a, b ,and d i believe. How do i solve c?
     
  14. Dec 9, 2011 #13
    What is necessary for acceleration to occur?
     
  15. Dec 9, 2011 #14
    change in velocity
     
  16. Dec 9, 2011 #15
    what causes a change in velocity?
     
  17. Dec 9, 2011 #16
    particle speeding up or slowing down
     
  18. Dec 9, 2011 #17
    That is what a change in velocity is. I asked what is the cause of that.
     
  19. Dec 9, 2011 #18
    change in position???
     
  20. Dec 9, 2011 #19
    Change in position is once again a result of something, not a cause.

    What is the source of all of this change. You mention change in velocity and position, but none of those happen on their own. What makes them happen?
     
  21. Dec 9, 2011 #20
    an outside force? in this instance the spring compressing and expanding causing the block to move?
     
  22. Dec 9, 2011 #21
    Yes! nice.

    So now that you know that a force has to cause this acceleration, when do you think (during the springs period of motion) the spring exerts the most force (therefore the most acceleration).
     
  23. Dec 9, 2011 #22
    when it is either compressed or expanded the most, the endpoints of the amplitude
     
  24. Dec 9, 2011 #23
    Correct. You got it.

    I must stress this, I'm sorry if I'm sounding like a broken record, but the energy calculations will show you a much higher level understanding of the problem. If you can understand that at maximum amplitude (the most force), that all of the energy is in the spring (the mass is at rest), then you now have automatic initial conditions for any equation of motion.

    The maximum speed occurs when all of the energy is in the mass (kinetic energy) which would mean when 0 energy is in the spring (equilibrium). Thinking in terms of energy for this kind of stuff is very useful.
     
  25. Dec 9, 2011 #24
    but that gives the maximum velocity not the maximum acceleration, correct?
     
  26. Dec 9, 2011 #25
    oh wait nvm!
     
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