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Homework Help: Just need Simple Harmonic Motion problem checked

  1. Feb 7, 2009 #1
    1. The problem statement, all variables and given/known data
    a 1kg block sliding a frozen surfaces collideds w/a horizontal spring w/a force constant of 2N/m. The block compresses the spring 4m fr. its rest position.
    a.how long does it take to fully colmpress the spring?b. what wa sthe speed of the block at the instant of the collision?c. when the springs is compressed 1.7m, how much force is the block on it?

    2. Relevant equations
    k=force constant A=4m x=1.7m
    T=2pi(sqroot (m/k)

    3. The attempt at a solution
    I used the T eq. to find the answr for a. which is 4.44s
    Then I took T and plugged in for vmax eq. and have the speed as 5.66m/s
    for c. I used the f/x=k equation and came up with F=3.4N

    is this the way I'm suppose to do it, it somehow seems too easy to be true? =x
  2. jcsd
  3. Feb 7, 2009 #2

    Tom Mattson

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    Your answer for a is wrong. And since you used your answer for a to answer b, b is wrong too. You used the equation of the period to find the time to fully compress the spring. But the period is the time required to execute a full cycle, not compress the spring once. You got c right.
  4. Feb 7, 2009 #3
    can you hint me cuz i cant find an another equation with T in it. =x sorry...
  5. Feb 7, 2009 #4
    He doesn't want you do use another equation, I reckon he wants you to understand something conceptually. T=period, and if you have an oscillating system and start at x=A, then the time it takes to go from there and back to the exact same position x=A is the period, T. So, a full cycle would be a full oscillation. If you go from x=A to x= -A, it would be ½T, meaning half an oscillation. The time you want to find is neither--your compression isn't a full cycle (which is what you've calculated) or ½ a cycle. You go from x=0 to x=A, which is how much of a full oscillation?
  6. Feb 7, 2009 #5
    1/T? sorry, but im a lil confused because a full oscillation is when the block hitted the spring and goes back to its original state right? so when it was compressed, it would be half of the compressed so its 1/4 of the T equation?
  7. Feb 7, 2009 #6
    Actually, as you go from rest to amplitude and back to rest again you only do ½ an oscillation--a full one is when you're at the same position with the velocity vector pointing the same way. When you go back to the rest position, the velocity is backwards (so to speak). So, the block would have to reach -A (meaning the full extension) and back again to the rest position to complete one oscillation. So, you have only a quarter of an oscillation. (Am I even making sense? it's late where I live and I'm tired.)

    But on the other hand, you are correct when you say it's 1/4 T.
  8. Feb 7, 2009 #7

    Tom Mattson

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    Yes, compressing the spring from its equilibrium position takes 1/4 of a period.
  9. Feb 7, 2009 #8
    thank you for taking your time to help me. im sorry for keeping you up.

    Imma try the problem again. and hopefully i'll get it right this time thank you.
  10. Feb 7, 2009 #9
    so i used (1/4)T=2pi(sqroot(m/k) and get 1.11s? then use this inthe velocity equation and get v= 22.64m/s?
    thanks for the help!
  11. Feb 7, 2009 #10
    is this correct?
  12. Feb 8, 2009 #11
    Yeah, I'd say that's correct.
  13. Feb 8, 2009 #12
    thank you so much for the help..i love this site... you all are so helpful and helps me understand the proble,m more...thank you all for the help...
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