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Plz help w/ Simple harmonics question

  1. Feb 5, 2008 #1
    [SOLVED] Plz help w/ Simple harmonics question

    I am doing simple harmonics and running into problems. The formulas given for simple harmonis in my book are f=[1 / 2(pi)]square_root (k /m) or
    f=[1 / 2(pi)]square_root (a /-x). The question reads:

    Car 1 has a spring-loaded rear bumper with a force constant of 840000 N/m. A second vehicle, car 2 (with a mass of 1500 kg), travels at a constant speed of 18 km/h, hitting car 1 in the rear bumper.

    a) calculate the kinetic energy for car 2.

    for this I used
    Ek = ½ mv^2
    Ek = ½ (1500 kg)(5 m/s)^2
    Ek = 1.9 x 10^4 J

    b) calculate the distance that car 1's bumper will compress if car 2 comes to a complete stop after striking it ?

    from the information given i can calculate the frequency but from there I can't figure the rest

    Also I have another harmonics question. Prove that the maximum speed (V{max}) of a mass on a spring is given by 2(pi)fA.

    where f = frequency and A =Amplitude? I'm guessing I have to rearrange one of formulas given at the top of the post. Thanks
     
  2. jcsd
  3. Feb 5, 2008 #2

    Hootenanny

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    Your answer to question (a) is correct. For question (b) I offer a hint: consider conservation of energy.
     
  4. Feb 5, 2008 #3
    the law of conservation states that energy cannot be created or destroyed. Would i calculate the potential energy of car 1 and add it to the kinetic energy of car 2?
     
  5. Feb 6, 2008 #4

    Hootenanny

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    Before the collision, car two has some kinetic energy. When car two collides with car one's bumper, it will compress the spring(s). After the collision, car two comes to a complete stop, assuming that car one's speed remains constant this means that all car two's kinetic energy has been converted to ______________.
     
  6. Feb 6, 2008 #5
    Potential energy ?
     
  7. Feb 6, 2008 #6
    so in this question Ek = Ep? from there where do i go to find the distance travelled by the spring?
     
  8. Feb 6, 2008 #7
    Potential energy ? so would Ek=Ep ? from there how do i find the distance that the spring travelled?
     
  9. Feb 6, 2008 #8

    Hootenanny

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    Correct :approve:
    Correct again :approve:
    What is the expression for the energy stored in a compressed spring?
     
    Last edited: Feb 6, 2008
  10. Feb 6, 2008 #9
    Et = 1/2mv^2 + 1/2kx^2
    I think i get it. I use the value above that I calculated for Ek, put that as Et in the formula then isolate to solve for x.
     
  11. Feb 7, 2008 #10

    Hootenanny

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    Correct :approve:
     
  12. Feb 7, 2008 #11
    originally i thought that was what i do. however i run into a problem when i do that. If Et = Ek and Ek= 1/2mv^2. If i look at the formula i should get
    1900 J = 1900 J + 1/2kx^2.
    1900 J - 1900 J = 1/2kx^2
    0 = 1/2 kx2
    i cant have a 0 divided by 1/2k
     
  13. Feb 8, 2008 #12

    Hootenanny

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    Note that initially v = 18 km/h therefore, your total energy will be equal to the kinetic energy of car 2 (1900J). After the collision v = 0 (the car has stopped), therefore the kinetic energy term disappears and you're equation becomes,

    1900 J = 1/2kx^2

    Do you follow?
     
    Last edited: Feb 8, 2008
  14. Feb 8, 2008 #13
    well explained I understand thank you very much
     
  15. Feb 8, 2008 #14

    Hootenanny

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    A pleasure :smile:
     
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