## 17.3: Applications of Second-Order Differential Equations

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 ## MATHEMATICA TUTORIAL, Part 1.4: Spring Problems

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 ## Section 5.1-2 Mass Spring Systems

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 You ask, we answer! Our team is dedicated to providing the best possible service to our customers. In order to determine what the math problem is, you will need to look at the given information and find the key details. Once you have found the key details, you will be able to work out what the problem is and how to solve it. Expert instructors will give you an answer in real-time

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## The Mass-Spring Oscillator

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 Get Assignment

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## Engineering Math

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

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## Applications of Second-Order Differential Equations

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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