A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially released from a p

Question

A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s. amplitude ft period s How many complete cycles will the mass have completed at the end of 6π seconds? cycles

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niczorrrr 5 months 2021-08-05T21:26:35+00:00 1 Answers 302 views 0

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    -1
    2021-08-05T21:28:19+00:00

    Answer:

    A) Amplitude = 1.803 ft ; Period = π/2 s

    B) Cycles in 6π s = 12 cycles

    Explanation:

    Let M be the mass attached and let K be the spring constant.

    Applying Newton’s second law to this system, we have;

    m(d²x/dt²) = -kx²

    When x(t) is the displacement from the equilibrium position, the equation can be given as;

    d²x/dt² + (k/m)x = 0 – – – – – (eq1)

    Let’s find the mass.

    W=mg

    W = 32lb and g = 32 ft/s²

    So, m = W/g = 32/32 = 1 lbft/s² = 1 slug

    Let’s find spring stiffness;

    From Hooke’s law, W = ks

    s = 2 ft. Thus, k = W/s = 32/2 = 16 lb/ft

    Now, putting the values of m and k into eq 1,we obtain;

    d²x/dt² + (16/1)x = 0

    d²x/dt² + 16x = 0

    Now, we know that the solution of x” + ω²x = 0 is given as;

    x(t) = C1 cosωt + C2 sinωt

    By inspection, ω² = 16. Thus, ω = 4

    So, the general solution is now;

    x(t) = C1 cos4t + C2 sin4t

    The initial conditions from the question are;

    x(0) = -1 ft and x'(0) = -6 ft/s

    At x(0) = -1

    -1 = C1 cos 0 + C2 sin 0

    -1 = C1

    Now, at x'(0) = 6

    x’ = -4C1sin4t + 4C2cos4t

    So,

    6 = -4C1sin 0 + 4C2cos0

    4C2 = – 6

    C2 = -6/4 = -3/2

    Thus, the equation of motion is;

    x(t) = – cos4t – (3/2)sin4t

    A) Amplitude is given as;

    A = √(C1² + C2²) = √(-1² + (-3/2)²)

    A = √(1 + 9/4) = √(13/4) = 1.803 ft

    Period is given as;

    T = 2π/ω = 2π/4 = π/2 s

    B) Cycles it will complete in 6π;

    = 6π/T = 6π/(π/2) = 12 cycles

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