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Finding postion, velocity and accelration as functions of time (SHM)

  1. Sep 5, 2011 #1
    1. The problem statement, all variables and given/known data

    A 1.00-kg glider attached to a spring with a force constant 16.0 N/m oscillates on a frictionless, horizontal air track. At t = 0, the glider is released from rest at x = -2.50 cm (that is, the spring is compressed by 2.50 cm).

    Find the position, velocity, and acceleration as functions of time. (Where position is in m, velocity is in m/s, acceleration is in m/s2, and t is in s. Use the following as necessary: t.)

    2. Relevant equations

    x(t)=Acos(ωt+ φ)

    v(t)= dx/dt

    a(t)=d^2x/dt^2

    ω=√(k/m)

    3. The attempt at a solution


    I know that the answer is x(t)= .025cos(4t+π) (then taking the derivatives respectively). What I can't figure out however is why φ=π. I thought if the particle is at its maximum position (x=A) at t=0, φ=0?
     
  2. jcsd
  3. Sep 5, 2011 #2

    vela

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    You're right that if t=0 and φ=0, the oscillator starts at x=A. The amplitude A is a positive number. Perhaps that's the bit of info you're missing. In this problem, you know the oscillator starts at the other extreme, x=-A, where A = 0.025 m.
     
  4. Sep 5, 2011 #3
    I copied the question word for word. It was the third question in the problem however. It asked me to find the period and the maximum velocity and acceleration. But I don't see how that would matter in the third question.
     
  5. Sep 5, 2011 #4

    vela

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    I have no idea what you're asking.
     
  6. Sep 5, 2011 #5
    I'm asking why φ=π in this question when I thought it should be zero because of the given information.
     
  7. Sep 5, 2011 #6

    HallsofIvy

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    My first thought, seeing a "glider" and a spring was that the glider was being lauched into the air by the spring! But that clearly is not the case- your "glider" is simply a mass attached to the spring so that it is pushed out and pulled back by the spring. Since the spring has spring constant 16 N/m, the force, at distance x from the neutral point is -16x Newton's. That is, you have [itex]mdx^2/dt^2= -16x[/itex] or [itex]dx^2/dt^2= -16x/m[/itex] which is just [itex]dx^2/dt^2= -16x[/itex] since the mass is 1 kg. The fact that the glider is " released from rest at x = -2.50 cm" means that x(0)= -0.0250 (because the spring constant is given in Newtons per meter, you want x in meters)and dx/dt(0)= 0.

    Since you apparently are given "x(t)=Acos(ωt+ φ)", just replace x in each of dx/dt= -16x, x(0)= -2.5, dx/dt(0)= 0. You will get three equations to solve for the three parameters, A, ω, and φ.

    (The condition [itex]d^2x/dt^2= -16x[/itex] must be true for all x. Comparing the right and left sides will tell you what ω is.)
     
  8. Sep 5, 2011 #7

    At first they asked me to find the period of this glider, so to find ω I just did 2π/T
     
  9. Sep 5, 2011 #8

    vela

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    I explained that earlier. When t=0, you get x(0) = A cos φ where x(0) = -0.025 m and A = +0.025 m. Solve for φ.
     
  10. Sep 5, 2011 #9
    Oh! i see it now. You'll get -1=cosφ making φ=π

    Thank you and sorry for overlooking what you told me I overlooked :uhh:
     
  11. Sep 5, 2011 #10

    gneill

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    Take advantage of the trig identity cos(θ + π) = -cos(θ).

    In this problem you want the initial displacement to be -A at time t = 0. By adding π to the argument of the cosine function you accomplish changing the sign.
     
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