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Math Help - Trig Set..

  1. #1
    WahedKhandabi
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    Trig Set..

    A buoy oscillates in simple harmonic motion Y=A cos cot as waves move past it. The buoy moves a total of 3.5 feet (vertically) from its low point to its high point. It returns to its high point every 10 seconds.
    a) Write an equation describing the motion of the buoy if it is at its high point at t = 0.
    b) Determine the velocity of the buoy as a function of t.
    c) Determine the times when the velocity of the buoy is greatest. What is the velocity at these times?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by WahedKhandabi View Post
    A buoy oscillates in simple harmonic motion Y=A cos cot as waves move past it. The buoy moves a total of 3.5 feet (vertically) from its low point to its high point. It returns to its high point every 10 seconds.
    a) Write an equation describing the motion of the buoy if it is at its high point at t = 0.
    b) Determine the velocity of the buoy as a function of t.
    c) Determine the times when the velocity of the buoy is greatest. What is the velocity at these times?
    No wonder you are having problems, your equation for SHO is messed up.

    y = A cos(omega*t + phi)
    where omega is the (angular) frequency and phi is the (angular) displacement. (Note: your book might use a different terminology for phi.)

    For all three of these we have the same frequency. So, the period T is 10 s. Thus omega = 2*pi/T = 0.62832 Hz (technically the unit is rad/s.)

    We also know that A = 3.5 ft because A is the magnitude of the max or min value of the wave.

    a) y(0) = 3.5 ft.
    y(t) = 3.5*cos(0.62832*t + phi)

    At t = 0 s:
    y(0) = 3.5*cos(0.62832*0 + phi) = 3.5

    Thus cos(phi) = 1 ==> phi = 0 rad.

    So y(t) = 3.5*cos(0.62832*t)

    b) v = dy/dt = -3.5*0.62832*sin(0.62832*t)
    v(t) = -2.1991*sin(0.62832*t) (ft/s)

    c) The velocity is greatest when dv/dt = 0, so:
    dv/dt = -2.1991*0.62832*cos(0.62832*t)

    0 = 7.69690*cos(0.62832*t) ==> cos(0.62832*t) = 0

    Thus
    0.62832*t = pi/2 + n*pi where n = 0, 1, 2, 3,....

    0.62832*t = pi/2 + n*pi ==> t = 2.5 + 5n (s)

    The velocity at these times are:
    v(max) = -2.1991*sin(0.62832*[2.5 + 5n])

    v(max) = -2.1991*sin(pi/2 + n*pi) <-- These are the exact values. The approximated values work nearly as well.

    Now, sin(A + B) = sin(A)cos(B) + sin(B)cos(A)
    So sin(pi/2 + n*pi) = sin(pi/2)cos(n*pi) + sin(n*pi)cos(pi/2)

    sin(pi/2) = 1
    cos(n*pi) = (-1)^{n}
    sin(n*pi) = 0 for all n
    cos(pi/2) = 0

    Thus sin(pi/2 + n*pi) = (-1)^{n}

    Finally:
    v(max) = -2.1991*(-1)^{n} = 2.1991*(-1)^{n+1} (ft/s)

    Probably your book was REALLY asking for the speed at these times which is a scalar, rather than the velocity which is a vector. The speed at these times is the magnitude of the velocity, and thus will be
    2.1991 ft/s.

    -Dan
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