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Math Help - simple harmonic motion

  1. #1
    Member sinewave85's Avatar
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    simple harmonic motion

    Ok, I am studying for a test and working through some practice problems on a handout from teacher. The problem is, I am getting a different answer than the one he gives for this application problem:

    "A tractor tire has a diameter of 6 feet and is revolving at a rate of 45 rpm. At t=0 a certain point is at height 0.
    A) Write an equation with phase shift 0 to describe the height of the point above the ground after 't' seconds.
    B) How high is the point above the ground after 28 seconds?"

    I get this:
    h = -3cos(t*8pi/3) + 3
    and at t = 28, h = 4.5

    but the answer given is this:
    h = -3cos(t*3pi/2)+3
    and at t = 28, h = 0

    The difference is the value for k, which I got by converting the period from revs per min (45/1) to rev per second (45/60 or 3/4), and then k = 2pi(4/3). The supplied answer inverts this, and I am wondering if this is a mistake or if I am looking at it incorrectly. Thanks.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by sinewave85 View Post
    Ok, I am studying for a test and working through some practice problems on a handout from teacher. The problem is, I am getting a different answer than the one he gives for this application problem:

    "A tractor tire has a diameter of 6 feet and is revolving at a rate of 45 rpm. At t=0 a certain point is at height 0.
    A) Write an equation with phase shift 0 to describe the height of the point above the ground after 't' seconds.
    B) How high is the point above the ground after 28 seconds?"

    I get this:
    h = -3cos(t*8pi/3) + 3
    and at t = 28, h = 4.5

    but the answer given is this:
    h = -3cos(t*3pi/2)+3
    and at t = 28, h = 0

    The difference is the value for k, which I got by converting the period from revs per min (45/1) to rev per second (45/60 or 3/4), and then k = 2pi(4/3). The supplied answer inverts this, and I am wondering if this is a mistake or if I am looking at it incorrectly. Thanks.
    Here is one way.

    The locus of the point in question relative to t in seconds is that of a sinusoidal curve---sine or cosine curve. Since the h is 0 at t=0, and the "spoke" or "center" of the tire is 3 ft above the ground, then the cosine curve is more applicable here because at h=0, the graph is at its lowest point.

    The equation then has the form
    h = A*cos[B(t-C)] +D --------------(i)
    where
    A = amplitude = 3 ft here.
    B = frequency in revolution/sec ----[Period = 2pi/B]
    C = horizontal shift or phase = 0 here
    D = vertical shift = 3 ft above the ground here.

    Hence, (i) becomes
    h = 3cos[B*t] +3 --------------(ii)

    Only B is missing.
    The tire moves at 45 rpm, at 45 revolutions per minute. Per second, that is
    (45 rev/min)(1min/60sec) = (45/60) rev/sec = 3rev /4sec.
    That means frequency B = 3rev/4sec. In radians, since 1 rev = 2pi radians,
    B = 3(2pi)/4sec = (3pi/2)rad/sec
    Thus, (ii) now becomes
    h = 3cos[(3pi/2)*t] +3 -------------------(iii)

    It is given that at t=0 sec, the h is 0 ft, so, in (iii),
    0 = 3cos[(3pi/2)*t] +3
    Without going any further, just by looking, it is obvious that for the righthand side to be equal to zero, "3cos[(3pi/2)t]" must be negative.
    Thererfore, finally, the (iii) becomes
    h = -3cos[(3pi/2)t] +3 ---------------------the equation, answer.

    ---------------------------
    B) "How high is the point above the ground after 28 seconds?"

    h = -3cos[(3pi/2)(28)] +3
    h = 0 -------------------------answer.
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