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Momentum, Impulse, Area under curve

  1. Sep 3, 2015 #1
    Δ≤ 1. The problem statement, all variables and given/known data

    Find final velocity.

    Knowns:
    m = 4kg
    Initial v = 0 m/s

    F = Asin(xt)
    [itex]F = (2000N)(sin(\frac{1000π}{sec}*t)[/itex] (0sec ≤ x ≤ .001sec)

    We need to find the impulse from a Force vs. Time graph.

    There is a preface to this problem that says if we work out the Force function, we will see that the contact period is approximately 1 millisecond.

    I don't see how that is possible. All I see here in regard to the Force function and the restriction is:


    [itex]((0sec ≤ \frac{1000π}{sec} ≤ .001sec)[/itex] which makes no sense to me...

    So is this question flawed from the beginning or am I completing not understanding what is being conveyed?

    2. Relevant equations

    F* Δt = mΔt

    3. The attempt at a solution

    Can not attempt yet. I don't understand the question to begin with.
     
  2. jcsd
  3. Sep 3, 2015 #2

    Orodruin

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    Is this the question exactly as stated, word by word? I suspect this is your interpretation of the question and as you say yourself, you do not understand the question. How do you then think we will understand the question based on your fragmented interpretation?
     
  4. Sep 3, 2015 #3

    SteamKing

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    It seems like there's a typo here. I think the range should be (0 s ≤ t ≤ 0.001 s)

    I think if you evaluate F = (2000 N) * sin (1000 π * t) on the interval (0 sec ≤ t ≤ 0.001 s), you'll see a force applied in a sinusoidal fashion.
     
  5. Sep 3, 2015 #4
    SteamKing,




    Why would the Professor in the video include the variable x in sin(xt) as part of the problem to begin with if we aren't to do anything with the x, other than to call it x=1000pi? What is the significance or relevance of x?

    In a previous problem, he used a quadratic equation for the Force function, but t was the only variable present in the Force function. Why two variables in this problem?
     
  6. Sep 3, 2015 #5

    haruspex

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    SteamKing is correct. The range specified for x in the image should be the range of t. The variable x here represents a constant, the value of which is given as 1000 pi s-1.
     
  7. Sep 3, 2015 #6

    SteamKing

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    Beats me. Professors also make mistakes (or the people making their videos do ...) I'm just offering what I think is a reasonable interpretation. You'll have to ask your professor about this to obtain a final resolution to the question.
     
  8. Sep 3, 2015 #7
    SteamKing,

    Thank you. I appreciate your clarification.

    Can any number (value) go into the sine function and make it sinusoidal? I just googled sinusoidal since I have not heard the term before. I understand that to be sinusoidal, a function must have the shape of a sine function. So If I have sin(1000/t), will it be sinusoidal? How about sin(1000t^2) or sin(2t/3)? In other words, what is it about F = (2000 N) * sin (1000 π * t) that conveys being of a sinusoidal fashion?
     
  9. Sep 3, 2015 #8

    SteamKing

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    The fact that the sine function is included in the definition of this force function.

    We learn about trig functions using expressions like sin (θ) or sin (t) or some such, but the bit inside the parentheses is not what makes something sinusoidal or not. It is the fact that the sine function itself is periodic and that it has a maximum and minimum amplitude, which repeat every so often because the sine is a periodic function.

    The general expression for a sine function is f(x) = A sin (bx + c) + d, where A is the amplitude, c/b is the phase shift, and d is the displacement (up or down along the y-axis).

    The argument of the sine, namely (bx + c), is measured in radians. The period is of this function is T = 2π / b

    http://www.teacherschoice.com.au/maths_library/functions/about_trigonometric_functions.htm

    by altering the values of A, b, c, or d, you can change the shape of the resulting function ...
     
  10. Sep 4, 2015 #9
    SteamKing,

    Two Questions:

    1) If a function is periodic but not sinusoidal, does it still contain a sin function?

    2) How does the Professor in the video determine

    [itex]F = (2000N)(sin(\frac{1000π}{sec}*t)[/itex] will exist within (0sec ≤ x ≤ .001sec) ?

    Is it a matter of finding the roots of the function? As in, let F(t) = 0

    [itex] 0 = (2000N)(sin(\frac{1000π}{sec}*t)[/itex]

    [itex]\frac{0}{2000N} = \frac{(2000N)(sin(\frac{1000π}{sec}*t)}{2000N}[/itex]

    [itex] 0 = (sin(\frac{1000π}{sec}*t)[/itex]

    [itex] sin^{-1}(0) = 1000π*t[/itex]

    [itex] 0 = 1000π*t[/itex]

    [itex] 0 = 1000π*t[/itex]

    t = 0.

    But what about the other root, t = .001?


    Thank you
     
  11. Sep 4, 2015 #10

    SteamKing

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    Not necessarily. The sine is just one example of a periodic function.

    If you plot the Force function for a series values in the range 0 ≤ t ≤ 0.001 s , it's easy to see that F(t) only takes positive values in this range.

    In other words, the professor used the properties of the sine function to construct this special force function, F(t).

    The sine function, f(θ) = sin (θ), has a period of 2π. sin (θ) is positive for the range 0 < θ < π and negative for the range π < θ < 2π. sin (nπ) = 0 for n = 0, 1, 2, ...

    If you specify that θ = 1000π * t, then when 0 < t < 0.001 s, this implies that 0 < θ < π.

    It also implies that sin (1000π * t) for 0 < t < 0.001s will produce the same values as sin (θ) for 0 < θ < π .

    This is all basic stuff from the study of trigonometric functions. Have you studied college trigonometry at all?
     
  12. Sep 5, 2015 #11
    [itex]F = (2000N)(sin(1000π*t)[/itex]

    Amplitude = 2000
    Period = 2pi =1000πt [itex] t = \frac{2}{1000}[/itex] = [itex] t = \frac{1}{500}[/itex]
    Phase Shift = (1000πt + 0) = [1000π(t +0/1000π)] = [1000π(t + 0)] = 0
    Vertical Shift = 0
    Count = (1/4)(1/500) = 1/2000
    Intervals = 0, 1/2000, 2/2000, 3/2000, 4/2000

    t = 0, 2000Sin(1000π(0)) = 0
    t = 1/2000, 2000Sin(1000π(1/2000)) = 2000Sin(π/2) = 2000
    t = 1/1000, 2000Sin(1000π(1/1000)) = 2000Sin(π) = 0
    t = 3/2000, 2000Sin(1000π(3/2000)) = 2000Sin(3π/2) = -2000
    t = 1/500, 2000Sin(1000π(1/500)) = 2000Sin(2π) = 0

    Along the Period of the function, there are 3 places where the range = 0. However, it is only between the first 2 zero points over which the sine function is positive:

    At
    t = 0 and t = 1/1000, or .001

    Thank you SteamKing.

    Regarding your question,
    Unfortunately I have not taken College Algebra or Trig, but I am working hard to learn the concepts as they arise within the Calculus and Physics material. It's a long story from involving simply not realizing I was interested in math and science until recently, to involving my school being somewhat lenient toward military circumstance I had at the time. Years later now, I have realized I love math and science and am working hard to learn the concepts as they arise within the Calculus and Physics material. I would gladly elaborate if you don't mind the digression. I'm willing to bet you've heard many situations of people slipping through the cracks of an educational system and then finding a real passion for math and science later on in life.

    Thank you for your guidance on this question, I now fully understand how to graph a trig function and will attempt the original velocity via Force vs. Time problem.
     
  13. Sep 5, 2015 #12

    SteamKing

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    I suspected that might be the case. My advice is to study college level trig and algebra before starting calculus. It'll save you a lot of trouble and headaches down the line...
     
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