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Simple harmonic motion

  1. Mar 19, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle moves with simple harmonic motion in a straight line with amplitude 0.05 m and the period 12 s.Find (a) the maximum speed , (b) the maximum acceleration of the particle.Write down the values of the constants P and Q in the equation
    x/m = P sin [Q)(t/s)]

    2. Relevant equations


    3. The attempt at a solution
    Well i tried to find the speed using A w sin w t
    but my problem is , how do i find the t ?
     
  2. jcsd
  3. Mar 19, 2015 #2

    DrClaude

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    Staff: Mentor

    ##t ## stays a variable. You should be able to restrict your analysis to one oscillation, as the motion will be periodic.
     
  4. Mar 19, 2015 #3
    if i use the equation A w sin wt
    i should replace t with what ?
     
  5. Mar 19, 2015 #4

    DrClaude

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    Staff: Mentor

    You are asked to calculate the maximum speed and acceleration. Do you know how to do that? For that, you need to be able to write position as a function of time. You then need to find the times that will maximize each of these values, for which, as I hinted, you can restrict yourself to one oscillation.

    By the way, the questions asks you to write it as x/m = P sin [Q(t/s)]. Use can use this instead of A and ω. Have you found P and Q?
     
  6. Mar 19, 2015 #5
    Its still not clear in my head :( Can you give me an example of how to write position as a function of time and where to take the max speed ?
    I am still trying to do the part (a) , when i finish i will do the remaining parts
     
  7. Mar 19, 2015 #6

    DrClaude

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    An example is given in the question:
    $$
    x = P \sin(Q t)
    $$
    You have position on the left-hand side, and a function of time on the right-hand side. Therefore, you can say that position is a function of time, that it varies with time.

    Finding P and Q is the first thing to do. I don't see how you can answer (a) or (b) without it. It may be the way the question is written that is confusing you. I read the last sentence of the problem as a hint, rather than a part (c) to be answered.
     
  8. Mar 19, 2015 #7
    Ah i see
    Yeah the question was confusing me xD

    Well
    x=Psin(Qt) you compare it to
    x=Asinwt

    p will be equal to 0.05
    and w(Q) will be π/6 right ?
     
  9. Mar 19, 2015 #8

    DrClaude

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    Correct.
     
  10. Mar 19, 2015 #9
    I have 1 question
    When do you know that you have to use either A w cos wt or A w sin wt ?
     
  11. Mar 19, 2015 #10

    DrClaude

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    It depends on the initial conditions (and your choice of ##t=0##). For a spring that is stretched at ##t=0## and then let go, the cosine will give you the right equation as you will get ##x(0) = A##.
     
  12. Mar 19, 2015 #11

    DrClaude

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    I should probably add that the general solution to the equation of motion of an harmonic oscillator is
    $$
    x(t) = A \cos (\omega t) + B \sin (\omega t)
    $$
    where ##A## and ##B## are set by the initial conditions (position and velocity at ##t=0##). It is left as an exercise :wink: to show that this can be rewritten as
    $$
    x(t) = C \sin (\omega t + \phi)
    $$
     
  13. Mar 19, 2015 #12
    Well if i choose X= A sin wt

    then v=dx/dt = A w cos wt which is what i want

    if i choose x = A cos wt it would be the inverse
    right ?
     
  14. Mar 19, 2015 #13

    DrClaude

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    Yes, and this corresponds, as I mentioned above, to the arbitrary choice of the zero of time.
     
  15. Mar 19, 2015 #14
    Well using v=dx/dt = A w cos wt i get 0.026 ms
    What formulae should i use for the acceleration ? Is it the same principle as for the speed ?
     
  16. Mar 19, 2015 #15

    DrClaude

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    Yes, same principle.
     
  17. Mar 19, 2015 #16
    so its A w^2 cos wt ?
     
  18. Mar 19, 2015 #17

    DrClaude

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    Yes, that's the equation for acceleration.

    [Edit: but with a minus sign. Thanks to BvU for spotting this.]
     
    Last edited: Mar 19, 2015
  19. Mar 19, 2015 #18
    Thanks very much for you help :)
     
  20. Mar 19, 2015 #19

    BvU

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    Gold Member

    Actually, there should appear a minus sign when differentiating a cosine.
    I hope that it is clear to you that when you start with
    the acceleration is given by a = dv/dt = -A w2 sin(wt)

    And conversely, when you start with
    X= A cos wt
    then v=dx/dt = -A w sin(wt)
    and a = dv/dt = -A w2 cos(wt)​
     
  21. Mar 19, 2015 #20
    well for v and a i got the correct answer by using the cos equation
    then when you derive the equation one should be cos and the other sin , for the v and a right?
    and thd minus sign is it necessary to put??
     
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