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, homework trouble, springs and ocilatory motion!

  1. Apr 6, 2007 #1
    plz help, homework trouble, springs and ocilatory motion!

    1. The problem statement, all variables and given/known data
    The position of a an air-track cart that is oscillating on a spring is given by (16.5cm)cos[(15.0s-1)t].
    At what value t after t=0 is the cart first located at x=4.6 cm?


    2. Relevant equations
    x=Acos[(angular freqency)(t)+(phase constant)] i think is the only one but i dont get the question right, maybe its my math.


    3. The attempt at a solution

    @=phase constant, w= angular frequency

    comparing x= 4.6cm=(16.5cm)cos[(15.0s-1)t] to:
    X=Acos[wt+@]
    inserting the information we get
    0.046m=0.165m*cos1.5(t)
    0.278=cos1.5t
    t=0.278s

    but incorrect.
     
  2. jcsd
  3. Apr 6, 2007 #2

    Mentz114

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    I don't see a phase constant in this -

    (16.5cm)cos[(15.0s-1)t]

    which has no constant in the [..] because it's all multiplied by t. Also the dimensions are wrong. Are you sure you haven't miscopied it ?
     
  4. Apr 6, 2007 #3
    the question itself is copied n pasted. the reason theres no phase constant is because apparently at that point phase constant = 0.. i dunno maybe im wrong about that but.. but the question is copied n pasted.
     
  5. Apr 6, 2007 #4

    Mentz114

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    Let me explain why I think it's wrong.

    Anything inside a cos() function must be dimensionless, like an angle. It cannot be in meters or Kg etc.
    Therefore the expression you've got

    15s-1 must be in units of 1/t so that when multiplied by t it becomes an angle. But 1 times t is in seconds and cannot be an angle.

    Usually its cos(wt+phi) as you've written, where w is 1/t and phi is already an angle.
     
  6. Apr 6, 2007 #5
    im confuzed right now :s
     
  7. Apr 6, 2007 #6

    Mentz114

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    OK, I've figured out your weird notation The 15s-1 is 15 radians per second. So your problem is straightforward. w is 15 and phase constant is zero.

    So use 15 in your calculation, not 1.5.
     
  8. Apr 6, 2007 #7
    ugh it still didnt work out right, i got like 0.0289 >_<
     
  9. Apr 6, 2007 #8
    then i got 0.289 when i tried again, but it still wasnt right
     
  10. Apr 6, 2007 #9

    Mentz114

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    I get

    t = [arccos( 0.278)]/15

    but I don't have arccos tables ...
     
  11. Apr 6, 2007 #10
    if i do that then i get 4.92 which apparently still aint right, unless im doing something wrong mathwise, i took the inverse cosine of 0.278 which was 73.9then divided it by 15. and got 4.92.
     
  12. Apr 6, 2007 #11

    Mentz114

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    The angle must be in radians, so 73.9 deg = 73.9/57.295 rad = 1.289 rad

    You'll get about .085 ...

    ( 57.295 = 180/pi )

    a complete cycle only takes about 0.42 secs ( 2*pi/15)
     
    Last edited: Apr 6, 2007
  13. Apr 6, 2007 #12
    WOO who i got it! thanks!
     
  14. Apr 6, 2007 #13

    Mentz114

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    Well done. Sorry I misunderstood the 15s-1.
     
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