Section 5.1 Notes - Fall 2002

# Section 5.1 Notes - Fall 2002 - A c 1 sin = and A c 2 cos...

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Differential Equations – Section 5.1 – Homework: 1, 3, 5, 11, 21, 23, 29 Hooke’s Law - ks F = , k = spring constant, s = amount of elongation, F =restoring force Newton’s Second Law Free Undamped Motion or Simple Harmonic Motion 0 2 2 2 = + x dt x d ϖ where m k / 2 = , 0 ) 0 ( x x = =the amount of initial displacement 1 ) 0 ( x x = = the initial velocity of the mass Solution: t c t c t x sin cos ) ( 2 1 + = The period of free vibrations is π / 2 = T and the frequency is 2 / / 1 = = T f . A maximum of x(t) is a positive displacement corresponding to the mass’s attaining a maximum distance below the equilibrium position, whereas a minimum of x9t) is a negative displacement corresponding to the mass’s attaining a maximum height above the equilibrium position. We refer to either case as an extreme displacement of the mass. The particular solution is the equation of motion. #2 #6 If you need to find the amplitude of motion, recall from trig: ( 29 φ + = + = t A t c t c t x sin sin cos ) ( 2 1 where 2 2 2 1 c c A + = and is the phase angle defined by

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Unformatted text preview: A c 1 sin = and A c 2 cos = . #8 Free Damped Motion – We will assume that the damping force is given by a constant multiple of dt dx . The motions will be modeled by 2 2 2 2 = + + x dt dx dt x d ϖ λ where m β = 2 , m k = 2 . Case I: 2 2-The system is said to be overdamped . The solution is + =----t e c e c e t x t t 2 2 2 2 2 1 ) ( . This equation represents a smooth an nonoscillatory motion. Case II: 2 2 =-The system is said to be critically damped . The solution is ( 29 t c c e t x t 2 1 ) ( + =-. Any slight decrease in the damping force would result in oscillatory motion. Case III: 2 2 <-The system is said to be underdamped . The solution is ( 29 t c t c e t x t 2 2 2 2 2 1 sin cos ) (-+-=-. #22...
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Section 5.1 Notes - Fall 2002 - A c 1 sin = and A c 2 cos...

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