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Simple harmonic motion velocity & time

  1. May 27, 2009 #1
    1. The problem statement, all variables and given/known data

    The velocity of an object in simple harmonic motion is given by v(t)= -(0.269 m/s)sin(15.0t + 2.00π), where t is in seconds. What is the first time after t=0.00 s at which the velocity is -0.150 m/s?


    2. Relevant equations
    already given in question


    3. The attempt at a solution
    so i tried finding the time by simply plugging -0.150 m/s in the equation v(t)= -(0.269 m/s)sin(15.0t + 2.00π) and solved for time i got a negative answer..when i tried to put that in it was wrong and so was the positive can someone tell me what i am doing wrong thankx :))
     
  2. jcsd
  3. May 27, 2009 #2

    CompuChip

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    Can you show us your work?
    Probably you are almost there. Note that the question explicitly asks for the first solution after t = 0. Do you understand what this has to do with the given equation which contains a sine?
     
  4. May 27, 2009 #3
    there are lots of different solutions to the inverse sine of a number, though sine is not directly symmetrical about the y axis.

    if you're still struggling, plot a rough graph of v against t.

    the question is asking for the first time V = -0.15 where t is posative.


    oh, and you're using radians?
     
  5. May 27, 2009 #4
    this is how i did it the long version

    v(t)= -(0.269 m/s)sin(15.0t + 2.00π)
    -0.15= -(0.269 m/s)sin(15.0t + 2.00π)
    -0.15/-0.269 = sin (15.0t + 2.00π)
    [sin-1(-0.15/-0.269)]-2.00π =15 t

    -5.691668/15 = t
    -0.379 = t
     
  6. May 27, 2009 #5
    this is how i did it the long version

    v(t)= -(0.269 m/s)sin(15.0t + 2.00π)
    -0.15= -(0.269 m/s)sin(15.0t + 2.00π)
    -0.15/-0.269 = sin (15.0t + 2.00π)
    [sin-1(-0.15/-0.269)]-2.00π =15 t

    -5.691668/15 = t
    -0.379 = t

    CompuChip can you clarify what you meant thanks ..and yes I am using radians all the way through
     
  7. May 27, 2009 #6

    CompuChip

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    What you did seems correct, except it doesn't give you the answer you want. Remember that the sine function is periodic, so it keeps repeating itself. You can easily see this if you make a graph. Suppose that the period is P (for sin(t) the period P = 2pi for example, while for sin(3t) it is P = 2pi/3). If you have a solution to your equation you can add or subtract any multiple of P (like 2P, 12P, 12356P) and get another solution.

    So first you should find out the period of sin(15.0t+2pi). Then add an appropriate multiple until you get something > 0.

    Sorry I don't have much time to explain thoroughly right now, is it clear what I'm talking about?
     
  8. May 27, 2009 #7
    i'm a bit confused so i know what my angular frequency is which is 15 rad/s and the period of that is 2pi(15)...... and then i get confused are you saying that i should change the period and make it 3pi(15) and plug that in to the eqaution to get the answer??
     
  9. May 27, 2009 #8

    diazona

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    You're right that the angular frequency is 15 rad/s, but the period is not 2π(15).
    Do you see the pattern in that? Once you do, what is the period of sin(15t)?
    Once you have found the period correctly, just follow that instruction. You can take the solution you found and add or subtract any multiple of P to get another solution; in fact, you can create an infinite list of solutions this way. (Don't try to write out the whole infinite list! ;-p Just figure out enough that you can see the pattern.) Now, out of all these solutions, what is the smallest one that is greater than 0?
     
  10. May 27, 2009 #9
    would the period be then 2pi/15??
     
  11. May 28, 2009 #10

    CompuChip

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    Yes, very good.

    By the way, did you also plot the graph as I suggested? (Using a graphical calculator or a computer program, for example)
     
  12. May 28, 2009 #11
    i have another question i was wondering where this period would go in the equation of velocity...also the nxt part has to do with position and was wondering if this looked right
    v(t)= -(0.269 m/s)sin(15.0t + 2.00π) changes to
    x(t)= (0.269/15) cos (15.0t+2.00π)??/
     
  13. May 28, 2009 #12

    CompuChip

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    That looks very much ok to me.

    If you doubt yourself, remember that the velocity is the derivative of x(t) with respect to time. So you can take the derivative and check that you do get v(t).
     
  14. May 28, 2009 #13
    also with my new period how do i use it with the velocity equation??
     
  15. May 28, 2009 #14
    hi ok so i finally understood what to do and was wondering if you could verify my answer since they removed the question from my homework application ...i got 0.0399 seconds
     
  16. May 29, 2009 #15

    CompuChip

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    In 4 decimals, I got 0.0394 but that's probably just a rounding difference (you might have rounded an intermediate result where I didn't).
    So you solved the question, very well!
     
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