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Mass on air track (SHM)

  1. Nov 2, 2007 #1
    1. The problem statement, all variables and given/known data
    A 23.0 kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position of the mass is defined to be at x=0. Somebody pushes the mass to the position x= 0.350 m, then lets go. The mass undergoes simple harmonic motion with a period of 4.70 s. What is the position of the mass 3.760 s after the mass is released?
    Consider the same mass and spring discussed in the previous problem. What is the magnitude of the maximum acceleration the mass undergoes during its motion?
    2. Relevant equations

    x = Acos(wt)
    2piw = T

    3. The attempt at a solution

    first I plugged in the given T to find w to use in the first equation, but now I have two unknowns, A and t. I know it isn't .350 because that's the amplitude for the time of the period, the amplitude I should use should be less than that, but I have no idea how to find that. Any ideas of what I'm doing wrong?
  2. jcsd
  3. Nov 2, 2007 #2


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    Staff: Mentor

    0.35 m is the maximum amplitude, A, with respect to the reference position.

    Using x = A cos [itex]\omega[/itex]t, for t = 0, x(t=0) = A = 0.35 m.

    4.70 s is the period T, and [itex]\omega=\frac{2\pi}{T}[/itex].

    So find the position at t = 3.760 s.
    Last edited: Nov 2, 2007
  4. Nov 2, 2007 #3
    I'm still not seeing what to do. I found omega for the period T to be 1.34 rad/s, and I get that .35 is the max amplitude, but I'm not seeing how that helps me find A at 3.76 s
  5. Nov 2, 2007 #4


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    Staff: Mentor

    Let's look at the equation x = A cos [itex]\omega[/itex]t,

    A = 0.35 m, [itex]\omega[/itex] = 1.34 rad/s, and we want to find x at t = 3.76 s

    x (t= 3.76 s) = 0.35 m cos (1.34 rad/s * 3.76 s), and remember that the argument of the cos is rad, not degrees.
  6. Nov 2, 2007 #5
    my calculator is in radian mode, I took .35 times cos(1.34*3.76), but I'm still not getting the right answer.
  7. Nov 2, 2007 #6
    anyone still willing to help me?
  8. Nov 2, 2007 #7
    please someone help!
  9. Nov 3, 2007 #8


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    Staff: Mentor

    Well the other part of this problem is where is the mass in terms of the phase (or phase angle) with respect to the periodic motion.

    At full deflection, in this case 0.35 m, the phase angle is [itex]\frac{\pi}{2}[/itex].

    The general form for simple harmonic motion is x = A sin ([itex]\omega[/itex]t + [itex]\theta[/itex]), where [itex]\omega[/itex] is the angular frequency and [itex]\theta[/itex] is the phase angle, and for [itex]\theta[/itex] = [itex]\frac{\pi}{2}[/itex], the form becomes

    x = A cos [itex]\omega[/itex]t, so at t = 0, x = A.

    The time of 3.76 s represents 0.8 of the period T = 4.70 s.

    So x (t = 3.76) = 0.35 m cos (2 pi * 0.8) = 0.35 m cos (1.6 pi) = 0.35 * 0.309 = 0.108 m.
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