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25 April, 23:14

If an undamped spring-mass system with a mass that weighs 6 lb and a spring constant 1 lb/in is suddenly set in motion at t = 0 by an external force of 2 cos 7t lb, determine the position of the mass at any time. (Use g = 32 ft/s2 for the acceleration due to gravity. Let u (t), measured positive downward, denote the displacement in feet of the mass from its equilibrium position at time t seconds.)

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  1. 25 April, 23:32
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    U (t) = (48/45) (cos 7t - cos 8t) seconds

    Explanation:

    Weight = 6lb

    K = 1lb/in or 12lb/ft

    U' (0) = 0 ft

    U'' (0) = 0 ft/s

    F (t) = 2 cos 7t lb

    We know that, w = mg

    Thus, m = w/g = 6/32 = 3/16 lb²/ft

    Thus, the initial value problem cam be written as;

    3u'' + 192u = 48 cos 7t

    Since, u' (0) = 0 ft and u'' (0) = 0 ft/s, the corresponding homogenous equation is;

    3u'' + 192u = 0

    With a characteristic of;

    3p² + 192p = 0

    So, p = ±√ (192/3) i = ±√64i = ±8i

    This is a pair of conjugate roots and thus the solution is;

    u (t) = c1cos 8t + c2sin8t

    And the particular solution can be written as;

    Y (t) = A cos 7t + B sin 7t

    Now, let's plug Y (t) into the initial value problem;

    3Y'' (t) + 192Y (t) = 2 cos 7t

    3 (-49Acos 7t - 49Bsin 7t) + 192 (A cos 7t + B sin 7t) = 48 cos 7t

    Thus;

    -147ACos 7t - 147BSin 7t + 192A cos 7t + 192B sin 7t = 48 cos 7t;

    45A Cos 7t + 45B Sin 7t = 48 cos 7t

    By inspection, A = 48/45 and B=0

    Hence, the particular solution is;

    Y (t) = (48/45) cos 7t

    The general solution will now be;

    U (t) = Uc (t) + Y (t)

    U (t) = c1 cos8t + c2 sin8t + (48/45) cos 7t

    Mow, let's find c1 and c2 from the conditions in the initial value problem.

    Thus;

    At u (0) = 0;

    0 = c1 cos 0 + c2 sin0 + (48/45) cos 0

    So, c1 = 48/45

    Also, at u' (0) = 0;

    0 = - 8c1 sin 0 + 8c2 cos 0 - (7) (48/45) sin 0

    8c2 = 0 and c2 = 0

    Thus, the general solution of u (t) is;

    U (t) = (48/45) (cos 7t - cos 8t) seconds
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