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Unformatted text preview: Ch 3.8: Mechanical & Electrical Vibrations Two important areas of application for second order linear equations with constant coefficients are in modeling mechanical and electrical oscillations. We will study the motion of a mass on a spring in detail. An understanding of the behavior of this simple system is the first step in investigation of more complex vibrating systems. Spring – Mass System Suppose a mass m hangs from vertical spring of original length l . The mass causes an elongation L of the spring. The force F G of gravity pulls mass down. This force has magnitude mg , where g is acceleration due to gravity. The force F S of spring stiffness pulls mass up. For small elongations L , this force is proportional to L . That is, F s = kL (Hooke’s Law). Since mass is in equilibrium, the forces balance each other: kL mg = Spring Model We will study motion of mass when it is acted on by an external force (forcing function) or is initially displaced. Let u ( t ) denote the displacement of the mass from its equilibrium position at time t , measured downward. Let f be the net force acting on mass. Newton’s 2 nd Law: In determining f , there are four separate forces to consider: Weight: w = mg (downward force) Spring force: F s =  k ( L+ u ) (up or down force, see next slide) Damping force: F d ( t ) =  γ u ′ ( t ) (up or down, see following slide) External force: F ( t ) (up or down force, see text) ) ( ) ( t f t u m = ′ ′ Spring Model: Spring Force Details The spring force F s acts to restore spring to natural position, and is proportional to L + u . If L + u > 0, then spring is extended and the spring force acts upward. In this case If L + u < 0, then spring is compressed a distance of  L + u , and the spring force acts downward. In this case In either case, ) ( u L k F s + = ( 29 [ ] ( 29 u L k u L k u L k F s + = + = + = ) ( u L k F s + = Spring Model: Damping Force Details The damping or resistive force F d acts in opposite direction as motion of mass. Can be complicated to model. F d may be due to air resistance, internal energy dissipation due to action of spring, friction between mass and guides, or a mechanical device (dashpot) imparting resistive force to mass. We keep it simple and assume F d is proportional to velocity. In particular, we find that If u ′ > 0, then u is increasing, so mass is moving downward. Thus F d acts upward and hence F d =  γ u ′ , where γ > 0. If u ′ < 0, then u is decreasing, so mass is moving upward. Thus F d acts downward and hence F d =  γ u ′ , γ > 0 . ), ( ) ( ′ = γ γ t u t F d Spring Model: Differential Equation Taking into account these forces, Newton’s Law becomes: Recalling that mg = kL , this equation reduces to where the constants m , γ , and k are positive....
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 Fall '09
 Csar
 Cos, Spring Force Details

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