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F=MA 2009 #23 (Power of a Spring)

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data
    #23
    https://aapt.org/physicsteam/2010/upload/2009_F-maSolutions.pdf

    2. Relevant equations
    P = Fv


    3. The attempt at a solution
    T = 2∏√(k/m)
    x = Asin(wt)
    v = Awcos(wt)
    a = -Aw2sin(wt)

    We are trying to maximize power, which is both dW/dt and Fv.
    Force is ma, given that mass is constant, we need to maximize acceleration.
    We need to maximize the velocity as well.
    Writing this out:
    Power = Fv
    = -A^2w^3sin(wt)cos(wt)
    To maximize sin(wt)cos(wt), we should let (wt) = pi/4
    wt = pi/4
    2pi/Tt = w
    2pi /Tt = pi/4
    2/Tt = 1/4
    1/T = 1/8t
    T = 8t

    This is utterly wrong. How can I do this?
     
  2. jcsd
  3. Jan 29, 2013 #2

    Simon Bridge

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    You know those questions cut-and-paste nicely right?

    23. A mass is attached to an ideal spring. At time t = 0 the spring is at its natural length and the mass is given an initial velocity; the period of the ensuing (one-dimensional) simple harmonic motion is T . At what time is the power delivered to the mass by the spring first a maximum?

    (A) t = 0
    (B) t = T /8
    (C) t = T /4
    (D) t = 3T /8 ← CORRECT
    (E) t = T /2

    It occurs to me that in SHM, the force has a special relationship with position, and so does the velocity.
     
  4. Jan 29, 2013 #3
    F = ma = md^2x/dt^2
    F = mdv/dt
    Power = d^2x/dt^2 * dx/dt
    Power = d^3x/dt^3
    Power = -Aw^3cos(wt)?

    Actually I think I violated rules of multiplication.
     
  5. Jan 29, 2013 #4

    cepheid

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    Your method involving F*v is FINE. You just didn't take into account the negative sign on the sin(wt)cos(wt), which changes the phase of the instantaneous power oscillation. Pi/4 turns out to be a minimum, and the maximum occurs at some multiple of this. To be safe, differentiate P(t), set it to zero, and figure out the difference between maxima and minima by looking at the second derivative. Use a trig identity to turn sin(wt)cos(wt) into something else that has no multiplication, so that you don't have to bother with product rule.

    Pi/4 gives you T/8, (I think you just messed up your algebra there).
     
  6. Jan 29, 2013 #5
    So lets call Aw^2 k for simplicity

    P is -ksin(2wt)/2
    P' is -k/4cos(2wt)
    We need to find where cos is -1 so P is positive and maximized. This occurs at 3pi\2

    2wt = 3pi\2
    w = 3pi\4t
    1/T= 2pi\
    T = 3\8t
     
  7. Jan 29, 2013 #6

    cepheid

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    A couple of things wrong here:

    1. The derivative P' is -kcos(2wt). You get a factor of 2 from your differentiation, not a factor of 1/2 coming out. You did your chain rule wrong.

    2. The extrema (maxima and minima) of P occur where P' is ZERO. To figure out which are maxima and which are minima, you need to look at P'' (...remember differential calculus?)
     
  8. Jan 29, 2013 #7
    You need the two because you only have one sinxcosx not 2 as required by the double angle formula
     
  9. Jan 29, 2013 #8

    cepheid

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    I'm not disputing the factor of 1/2 in P. I'm disputing the factor of 1/4 in P'.

    (1/2) * 2 = 1, NOT 1/4.

    Edit: It doesn't matter too much, because it doesn't affect what wt has to be in order to maximize the function.
     
  10. Jan 30, 2013 #9

    Simon Bridge

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    The long way would be needed to gain understanding - isn't there a shorter approach by noticing that all the times are multiples of T/8 (multiples of pi/4 phase)? Those seem like points that are easy to find on a sinusoidal graph.

    Arn't we looking for where the kinetic energy vs time graph is changing the fastest?
    Or did I miss something?
     
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