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Simple harmonic motion: time for maximum speed

  1. Jan 19, 2012 #1
    1. The problem statement, all variables and given/known data

    A particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the origin, at t = 0 and moves to the right. The amplitude of its motion is 3.20 cm, and the frequency is 1.40 Hz.

    Determine the maximum speed of the particle.

    Determine the earliest time (t > 0) at which the particle has this speed

    2. Relevant equations
    V max = ωA

    f = 1/T

    T = 1/f

    ω = 2∏/T

    x(t) = Acos(ωt + ∅)

    max. acceleration = ω2A

    3. The attempt at a solution
    I only have a problem finding the time at which the particle reaches the max speed, don't know where to start with that at all, but:

    f = 1/T

    T = 1/1.4

    T = 0.714 s

    ω = 2∏/ T

    ω = 8.7999 rad / s

    Vmax = ωA

    Vmax = 0.28 m/s

    Vmax = 28.1 cm/s


    Where do I start with finding the time for the maximum speed?

    Edit: Okay someone told me I should get the equation as a function of time for it:

    which is X(t) = Acos(ωt + ∅ ) ??

    so i got x(t) = 0.032cos(8.8t + ∅ )

    derivative is, for v,

    v = -wAsin(wt + ∅)
     
    Last edited: Jan 19, 2012
  2. jcsd
  3. Jan 20, 2012 #2

    ehild

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    To make your X(t) function complete, use the information that the motion starts from equilibrium, that is, at X(t=0) = 0.
    At what position is the speed maximum?

    ehild
     
  4. Jan 20, 2012 #3
    0, 2∏ ?

    I honestly do not know what you are trying to imply.

    But the position where the speed is maximum is I guess at the beginning of the motion, 0, 2∏ and then over and over again.

    x(t) = Acos(ωt + ∅)

    x(t) = 0.032cos(8.8t + ∅ )

    0 = (0.032)cos[(8.8)(0) + ∅) ?????



    Vmax = -wAsin(wt + ∅)

    0.281 = (-0.282)sin(8.8t + ∅)

    Solve for t ?

    0.281/ (-0.282) = sin(8.8t + ∅)

    -0.996 = sin(8.8t + ∅)

    sin-1(-0.996) = (8.8t + ∅)

    -1.486 = 8.8t + ∅

    I think ∅ is 2∏

    -1.486 = 8.8t + 2∏

    t = -0.88 but this can't be right at all


    please help, thank you

    by the way i have to do the same thing for acceleration
     
  5. Jan 21, 2012 #4

    ehild

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    Yes, the speed is maximum at t=0. But that happens twice during a cycle. Do not forget that speed is magnitude of velocity. And you need to give the earliest time after the start when the maximum speed is reached. 2pi is not time.

    What is ∅ then?

    ehild
     
  6. Jan 21, 2012 #5
    I meant the position not time, it seems easier to think of it that way...


    continuing from what you said:

    0 = (0.032)cos[(8.8)(0) + ∅)

    so im solving for ∅? and cos(whatever) should equal 0 ?

    in that case:

    cos-1(0) = (∏/2) ???

    so ∅ = ∏/2 ????


    still do not really know what i am supposed to do or what you are trying to imply, but thank you, so far
     
    Last edited: Jan 21, 2012
  7. Jan 21, 2012 #6

    ehild

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    What should be the value of ∅ so as x(0)=0? It is not pi.

    The speed is maximum at position x=0, but x can not be 2pi.
    What is the time when the particle returns next time to x=0?


    ehild
     
  8. Jan 21, 2012 #7
    ugh

    i said it was ∏/2 not ∏ or am i missing something ???

    cos 90 deg = 0

    90 deg -> ∏ / 2 rad

    the time when the particle returns to x = 0 is the period... T = 0.714 s

    but it is not what i am looking for... and i dont really see how it helps that much

    what am i doing wrong??
     
  9. Jan 21, 2012 #8

    ehild

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    Sorry, I misread it. Well, it is pi/2 or -pi/2 if x(0)=0. But it is said that the particle moves to the right. If moving to the right means positive velocity ∅=-pi/2.

    Or it would be simpler to say that x(t) = 0.032sin(ωt).

    No. How many times is x=0 during a period?

    What is you are looking for? The question is:

    The particle has maximum speed at x=0. When is the particle next time at x=0?


    ehild
     
  10. Jan 21, 2012 #9

    gneill

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    Here's a diagram that may help to think about the problem:

    attachment.php?attachmentid=42920&stc=1&d=1327165036.gif
     

    Attached Files:

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  11. Jan 22, 2012 #10
    I finally got it right... t = 0.357s

    It is just the period time in half! I might still be missing a point though;

    It seems though that the equation didn't help much at all.

    I had NO IDEA why you guys were trying to set the position to x=0 in the equation for it, you just could have said sooner

    t = 0.357 when x = 0 because it is half of the period when x = 0 AGAIN. Thanks.

    and from the graph I would say the max acc begins at the amplitude / height of it.

    So instead of x = 0 it would be x = ∏/2 or x = 3∏/2 when trying to find the time to reach position of max acceleration right?
     
  12. Jan 22, 2012 #11

    ehild

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    The equation should have been x(t)=0.032 sin(2.8π t).

    v(t)=0.032(2.8π) cos(2.8π t).

    The speed (magnitude of v) is highest when the value of the cosine is ±1, that is, when the phase is 2.8πt = 0, π, 2π, ...For the soonest after zero t=π/(2.8π)=0.357

    ∏/2 or 3∏/2 is not X, but the phase when the acceleration is of maximum magnitude. X= ±0.032 m there.

    ehild
     
  13. Jan 22, 2012 #12
    umm... why?

    how??

    The equation i had in my first post was wrong though...

    i have now x(t) = 3.2 cos(8.8t - pi/2)



    i thought it was always x(t) = Acos(ωt + ∅) ??


    seems like i just got lucky by dividing the period time in half then... ugh

    yeah thanks I forgot to see that.


    thanks a lot, you are good and smart :cool:
     
    Last edited: Jan 22, 2012
  14. Jan 22, 2012 #13

    ehild

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    It can be Asin(wt+ψ ) as well. Both are SHM, and choosing appropriate phase constant, they are the same. Asin(ωt+π/2 )=Acos(ωt), Acos(ωt-π/2)=Asin(ωt).

    That dividing by half was OK, as the motion started with highest speed and there are two times when the speed is maximum during a cycle.

    Next time draw a diagram before you do any calculations. Something like the one gneill posted. Or make it plot out by wolframalpha.

    ehild
     
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