Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

JenniferSmart1 Group Title

Why is kx=25N instead of -kx=25N?

  • one year ago
  • one year ago

  • This Question is Closed
  1. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    • one year ago
  2. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @Outkast3r09

    • one year ago
  3. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    shouldn't F=-kx?

    • one year ago
  4. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @CliffSedge ?

    • one year ago
  5. Outkast3r09 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Ahh this my physics teacher went over

    • one year ago
  6. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    cool! Now you can explain it to me :)

    • one year ago
  7. Outkast3r09 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    the negative explains how that force is acting upon the spring

    • one year ago
  8. Outkast3r09 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    it depends on the what you're trying to find

    • one year ago
  9. Outkast3r09 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    if you're trying to find the force the spring is exerting on the mass or the mass exerting on the spring

    • one year ago
  10. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Hmmm let's see|dw:1353379482698:dw|

    • one year ago
  11. Outkast3r09 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1353379572781:dw|

    • one year ago
  12. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    well I guess we're stretching to a certain length, so the first figure?

    • one year ago
  13. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I mean the first figure you drew

    • one year ago
  14. Outkast3r09 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    the figures are the same

    • one year ago
  15. Outkast3r09 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    it depends on whether you're trying to find the length of the spring or the position of the mass

    • one year ago
  16. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Uhm I think we want my figure because In your figure we have to consider the gravitational force and the restoring force. I don't think they want to consider the gravitational force

    • one year ago
  17. Outkast3r09 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    if it asked abouthe spring itself it'd be negative

    • one year ago
  18. Outkast3r09 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    there is two ways this question can be answered 1) How long did the spring stretch from initial 2) What is the position of the mass after initial. They're the same thing

    • one year ago
  19. Outkast3r09 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    so since it's asking for position of the mass it's this|dw:1353379915405:dw|

    • one year ago
  20. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    why is it positive? because it is going in the direction that we designated as positive?

    • one year ago
  21. Outkast3r09 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yes

    • one year ago
  22. Outkast3r09 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    technically this question can have 2 answers + answer and a - answer depending on how you set up your axis

    • one year ago
  23. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    OH I see! |dw:1353380068540:dw| Like so?

    • one year ago
  24. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Now I can say that F=kx

    • one year ago
  25. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    right?

    • one year ago
  26. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    no the other way around, I should probably make the restoring force positive...

    • one year ago
  27. Outkast3r09 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1353380229463:dw|

    • one year ago
  28. Outkast3r09 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah the picture they used was probably a free falling mass

    • one year ago
  29. Outkast3r09 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    and saying that the sum of the force of gravity + any other force = F

    • one year ago
  30. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Yeah that makes more sense. so they probably had up as positive and down as negative.

    • one year ago
  31. Outkast3r09 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yes but it really just depends on what you define it... if your teacher marks you wrong and your diagram says otherwise just clear it up with him

    • one year ago
  32. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yep

    • one year ago
  33. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    so why is k(0.2) why 0.2? How did they come up with x=0.2?

    • one year ago
  34. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    never mind

    • one year ago
  35. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    LOL

    • one year ago
  36. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    that's the displacement haha

    • one year ago
  37. Outkast3r09 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yes the displacement will still be posive

    • one year ago
  38. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    hmm interesting: (this the DE part of my calc II book) \[2\frac{d^2x}{dt^2}+128x=0 \text{ has the solution }x(t)=c_1cos8t+c_2sin8t \] yeah that's easier that doing it manually....

    • one year ago
  39. CliffSedge Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    It's convention to designate potential energy as negative. The reaction force of the spring is opposite the positive applied force.

    • one year ago
  40. CliffSedge Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    More clearly: The negative sign of F=-kx means that the spring force is opposite to the applied force on the spring.

    • one year ago
  41. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    potential energy as positive...the restorative force?

    • one year ago
  42. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes that makes sense

    • one year ago
  43. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I had to reread the last sentence you wrote several times, but it makes sense now

    • one year ago
  44. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    but where did they get the 8 in sin and cos 8t

    • one year ago
  45. CliffSedge Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    As others mentioned above, it's a sign convention. You define what direction you want to be positive and anything opposite to that is negative. The norm, though is to treat spring force as negative/opposite to any force acting *on* the spring.

    • one year ago
  46. CliffSedge Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    The 8 is based on the period of the spring - that is determined by the spring constant, k, and the attached mass.

    • one year ago
  47. CliffSedge Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    \(\large T=2\pi \sqrt{\frac{m}{k}}\) T^-1 is the frequency of oscillation.

    • one year ago
  48. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yep, sounds good. I googled it LOL \[T=2\pi \sqrt{\frac mk}\]

    • one year ago
  49. CliffSedge Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    The "8" in the cosine function usually goes by the name, omega, \(\large \omega\). Might want to look that up too.

    • one year ago
  50. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    so the formula is \[x(t)=c_1cosTt+c_2sinTt\]

    • one year ago
  51. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    so T is \[\omega\]

    • one year ago
  52. CliffSedge Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    \(\large . . . cos(\omega t) . . \) No, ω is related to T.

    • one year ago
  53. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh \[\omega=\sqrt{\frac km}\]

    • one year ago
  54. CliffSedge Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Right!

    • one year ago
  55. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[x(t)=c_1cos\omega t+c_2sin\omega t\]

    • one year ago
  56. CliffSedge Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, you can either derive that using the Diff.Eq. or look it up on a physics formula sheet. (Depends on if you are more mathematician or engineer ;-) )

    • one year ago
  57. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    leaning more towards engineering....physics formula sheet :P

    • one year ago
  58. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    omega doesn't seem to have units

    • one year ago
  59. CliffSedge Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    That would be my choice too, but know that the way the physicists got those formulas in the first place was to solve the DEs (or rather to get their math dept. grad students and TAs to do the solving for them..)

    • one year ago
  60. CliffSedge Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    omega should have units of s^-1, it's a frequency.

    • one year ago
  61. CliffSedge Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Or .. is at least based on a frequency, something in the derivation might make the units cancel, but I'm pretty sure it's just the reciprocal of the time period, T.

    • one year ago
  62. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    radians per second

    • one year ago
  63. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    according to my friend wiki

    • one year ago
  64. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    so s^-1 should be sufficient, i can leave the radians part out in my notation?

    • one year ago
  65. CliffSedge Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, because radians are dimensionless (it's derived from length ÷ length), that's what the 2π conversion factor is in there for.

    • one year ago
  66. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh I see

    • one year ago
  67. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    why is c_2 zero? shouldn't it be ...so when we say\[ x(0)=8c_2 cos8t\] does that mean \[0=8c_2 cos8t=8c_2cos(0)=8c_2(1)=c_2\] yep that seems right LOL

    • one year ago
  68. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Thanks everyone!!!!

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.