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Homework Help: Simple harmonic motion

  1. Sep 20, 2010 #1
    1. The problem statement, all variables and given/known data

    A machaine part moves in a straight line with SHM. At positons 15mm and 20mm from the center of oscillation it has velocitys 2 m/s and 1 m/s respectively. find the amplitude and period of the motion. and the shortest time taken for the part to travel between these 2 positions.

    2. Relevant equations

    angular v= velocity * radius
    A=Xcos(Angular v*time)
    2pi*frequency= angular v
    Period= 1/frequency
    3. The attempt at a solution

    i have no idea.. its difficult for me to think of anything to help me find angular v ot frequency.
     
  2. jcsd
  3. Sep 27, 2010 #2

    nvn

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    Science Advisor
    Homework Helper

    Thk-MechEng: A couple of your relevant equations were incorrect.
    [strike]angular velocity = v*radius[/strike] w = angular velocity (rad/s) = 2*pi*f
    [strike]A = Xcos(angular velocity*time)[/strike] x = A*sin(w*t)
    v = dx/dt = A*w*cos(w*t)
    period = 1/f​

    Given:
    (1) 0.015 m = A*sin(w*t1).
    (2) 2 m/s = A*w*cos(w*t1).

    (3) 0.020 m = A*sin(w*t2).
    (4) 1 m/s = A*w*cos(w*t2).​

    Hint 1: Multiplying both sides of eqs. 1 and 3 by w, then squaring both sides of all four equations, gives,
    (1) [(0.015 m)*w]^2 = [(A*w)^2]*sin(w*t1)^2.
    (2) (2 m/s)^2 = [(A*w)^2]*cos(w*t1)^2.

    (3) [(0.020 m)*w]^2 = [(A*w)^2]*sin(w*t2)^2.
    (4) (1 m/s)^2 = [(A*w)^2]*cos(w*t2)^2.​

    Hint 2: Adding eqs. 1 and 2 together, and adding eqs. 3 and 4 together, gives,
    (1+2) [(0.015 m)*w]^2 + (2 m/s)^2 = [(A*w)^2]*[sin(w*t1)^2 + cos(w*t1)^2].
    (3+4) [(0.020 m)*w]^2 + (1 m/s)^2 = [(A*w)^2]*[sin(w*t2)^2 + cos(w*t2)^2].​

    Hint 3: Subtract eq. 3+4 from eq. 1+2. After you do that, see if you can now solve for w. After that, see if you can figure out how to solve for amplitude A.
     
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