pre algebra

A 0.25 kg mass is horizontally attached to a spring with a stiffness 4 N/m. The damping
constant b for the system is 1 N-sec/m. If the mass is displaced 0.5 m to the left and given an
initial velocity of 1 m/sec to the left, find the equation of motion. What is the maximum
displacement that the mass will attain?

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  1. Differential equations, initial value problem.

    The general equation of motion is:
    mx"+Bx'+kx=f(t), where the independent variable is t, and the displacement x is the dependent variable.

    In this case, external force f(t)=0, so
    mx"+Bx'+kx=0
    substitute
    m=0.25,
    k=4,
    B=1
    we have a differential equation of motion with constant coefficients:
    0.25x"+x'+4x=0
    Divide by the leading coefficient:
    x"+4x'+16x=0
    Auxiliary equation:
    m²+4m+16=0
    m1=-2+2√3, m2=-2-2√3
    α=-2, β=2√3
    So the solution for x is:
    x=e-2t(C1*cos(βt)+C2*sin(βt))
    Since x(0)=-0.5, we have
    -0.5=1*(C1+C2*0)
    or C1=-0.5

    Find
    x'
    =dx/dt
    =a*%e^(a*t)*(sin(b*t)*C2+cos(b*t)*C1)+%e^(a*t)*(b*cos(b*t)*C2-b*sin(b*t)*C1)
    substitute
    x'(0)=-1 to get
    -1=αC1+βC2
    -1=-2C1+2√(3)C2
    C2=(-1+2(-0.5))/2√(3)
    =-√(3)/3
    Therefore the equation of motion is:
    x(t)=e-2t(-0.5cos(βt)-(√(3)/3)sin(βt))

    Check my arithmetic.

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