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Spring Problems II Free Vibrations With Damping In this section we consider the motion of an object in a spring–mass system with damping. We start with unforced motion, so the equation of motion is my″ +cy +ky =0. (1) Now

The spring will deflect a distance h, where h is found from the relationship Mg = hK. Mg = hK is the statement that for static equilibrium, the weight must equal the force in the

y = e − ct / 2m(c1cosω1t + c2sinω1t). By the method used in Section 6.1 to derive the amplitude–phase form of the displacement of an

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If the spring is stretched then ΔL > 0 and Fs > 0, so the spring force is upward, while if the spring is compressed then ΔL < 0 and Fs < 0, so

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If the object is displaced y units from its equilibrium position, the total change in the length of the spring is ΔL =Δl−y, so Hooke’s law implies that Fs =kΔL=kΔl−ky. Substituting this into ( eq:6.1.1) yields my″ = −mg−cy +kΔL−ky+F. Since mg

Homework Equations The Attempt at a Solution I solve for k and get 64, and solve for the mass and get 32/64, so my differential equation is 0.5y'' + 2y' + 64y = 0, I solve for r and

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