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, homework trouble, springs and ocilatory motion!

  • Thread starter jaysea
  • Start date
  • #1
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plz help, homework trouble, springs and ocilatory motion!

Homework Statement


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?


Homework 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.


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.
 

Answers and Replies

  • #2
5,428
291
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 ?
 
  • #3
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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.
 
  • #4
5,428
291
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.
 
  • #5
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im confuzed right now :s
 
  • #6
5,428
291
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.
 
  • #7
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ugh it still didnt work out right, i got like 0.0289 >_<
 
  • #8
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then i got 0.289 when i tried again, but it still wasnt right
 
  • #9
5,428
291
I get

t = [arccos( 0.278)]/15

but I don't have arccos tables ...
 
  • #10
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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.
 
  • #11
5,428
291
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:
  • #12
7
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WOO who i got it! thanks!
 
  • #13
5,428
291
Well done. Sorry I misunderstood the 15s-1.
 

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