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

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.

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  1. A mass weighing 32 pounds stretches a spring 2 feet.

    (a) 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.

    (b) How many complete cycles will the mass have completed at the end of 4 seconds?

    Answer:

    [tex]A = 1.803 ft[/tex]

    Period = [tex]\frac{\pi}{2}[/tex] seconds

    8 cycles

    Explanation:

    A mass weighing 32 pounds stretches a spring 2 feet;

    it implies that the mass (m) = [tex]\frac{w}{g}[/tex]

    m= [tex]\frac{32}{32}[/tex]

    = 1 slug

    Also from Hooke’s Law

    2 k = 32

    k = [tex]\frac{32}{2}[/tex]

    k = 16 lb/ft

    Using the function:

    [tex]\frac{d^2x}{dt} = – 16x\\\frac{d^2x}{dt} + 16x =0[/tex]

    [tex]x(0) = -1[/tex]        (because of the initial position being above the equilibrium position)

    [tex]x(0) = -6[/tex]          ( as a result of upward velocity)

    NOW, we have:

    [tex]x(t)=c_1cos4t+c_2sin4t\\x^{‘}(t) = 4(-c_1sin4t+c_2cos4t)[/tex]

    However;

    [tex]x(0) = -1[/tex] means

    [tex]-1 =c_1\\c_1 = -1[/tex]

    [tex]x(0) =-6[/tex] also implies that:

    [tex]-6 =4(c_2)\\c_2 = – \frac{6}{4}[/tex]

    [tex]c_2 = -\frac{3}{2}[/tex]

    Hence, [tex]x(t) =-cos4t-\frac{3}{2} sin 4t[/tex]

    [tex]A = \sqrt{C_1^2+C_2^2}[/tex]

    [tex]A = \sqrt{(-1)^2+(\frac{3}{2})^2 }[/tex]

    [tex]A=\sqrt{\frac{13}{4} }[/tex]

    [tex]A= \frac{1}{2}\sqrt{13}[/tex]

    [tex]A = 1.803 ft[/tex]

    Period can be calculated as follows:

    = [tex]\frac{2 \pi}{4}[/tex]

    = [tex]\frac{\pi}{2}[/tex] seconds

    How many complete cycles will the mass have completed at the end of 4 seconds?

    At the end of 4 seconds, we have:

    [tex]x* \frac{\pi}{2} = 4 \pi[/tex]

    [tex]x \pi = 8 \pi[/tex]

    [tex]x=8[/tex] cycles

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