A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 2

Question

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 2 inches above the equilibrium position. Find the equation of motion. (Use g

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Thiên Hương 3 years 2021-08-05T18:29:51+00:00 1 Answers 1498 views 1

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  1. ngochoa
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    2021-08-05T18:31:02+00:00
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    Answer:

    the equation of motion is

    x(t) =-\frac{1}{6} \cos4\sqrt{6} t

    Explanation:

    Given that,

    The weight attached to the spring is 24pounds

    Acceleration due to gravity is 32ft/s²

    Assume x is the string length, 4inches

    convert the length inches to to feet = 4/12 = 1/3feet

    From Hookes law , we calculate the spring constant k

    k = W / x

    k = 24 / (1/3)

    k = 24 / 0.33

    k = 72lb/ft

    If the mass is displace from its equilibrium position by amount x

    the differential equation is

    m\frac{d^2x}{dt^2} + kx=0\\\\\frac{3}{4 } \frac{d^2x}{dt^2}+72x=0\\

    \frac{d^2x}{dt^2} +96x=0

    Auxiliary equation is

    m^2+96=0\\m=\sqrt{-96} \\m=\_ ^+4\sqrt{6}

    Thus the solution is,

    x(t) = c_1cos4\sqrt{6t} +c_2sin\sqrt{6t}

    x'(t) =-4\sqrt{6c_1} sin4\sqrt{6t} +c_24\sqrt{6} cos4\sqrt{6t}

    The mass is release from rest

    x'(0) = 0

    -4\sqrt{6c_1 } \sin4\sqrt{6} (0)+c_24\sqrt{6} \cos4\sqrt{6} (0)=0\\c_24\sqrt{6} =0\\\\c_2=0

    Therefore

    x(t) = c₁ cos4 √6t

    x(0) = -2inches

    c₁ cos4 √6(0) = 2/12feet

    c₁= 1/6feet

    There fore, the equation of motion is

    x(t) =-\frac{1}{6} \cos4\sqrt{6} t

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