# lec9-2 - Linear second-order differential equations with...

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Unformatted text preview: Linear second-order differential equations with constant coefficients and nonzero right-hand side • We return to the damped, driven simple harmonic oscillator d 2 y dt 2 + 2 b dy dt + ω 2 y = F sin ω t • We note that this differential equation is linear • We call y c the general solution which solves the homogeneous equation d 2 y c dt 2 + 2 b dy c dt + ω 2 y c = 0 • We call y p the particular solution which solves the inhomogeneous equation d 2 y p dt 2 + 2 b dy p dt + ω 2 y p = F sin ω t Linear equations with nonzero right-hand side continued • Because the differential equation is linear, in general y ( t ) = y c ( t ) + y p ( t ) • For the damped, driven simple-harmonic oscillator, with F ( t ) = F sin ω t driving force, y ( t ) = y c ( t ) + F q ( ω 2- ω 2 ) 2 + 4 b 2 ω 2 sin ( ω t- φ ) • The y c depends on whether the system is underdamped, overdamped, or critically damped • To find a solution to a given situation, we need to know the position and velocity of the oscillator at some time (eg t = 0), or alternately the position at two different times Example: Damped, driven simple harmonic oscillator • Example: At t = 0 we have y ( t = 0) = 0 and dy dt | t =0 = 0 • The system is underdamped, so that b < ω • The equation of motion is given by d 2 y p dt 2 + 2 b dy p dt + ω 2 y p = F sin ω t • The solution is y ( t ) = y c ( t ) + y p ( t ) y c ( t ) = ce- bt sin ( β t + γ ) • Here β = q ω 2- b 2 y p ( t ) = F q ( ω 2- ω 2 ) 2 + 4 b 2 ω 2 sin ( ω t- φ ) tan φ = 2 b ω ω 2- ω 2 Example continued • So we have expressions for y ( t ) and dy dt , y ( t ) = ce- bt sin ( β t + γ ) + F q ( ω 2- ω 2 ) 2 + 4 b 2 ω 2 sin ( ω t- φ ) dy dt =- ce- bt [ b sin ( β t + γ )- β cos ( β t + γ )]+ F ω q ( ω 2- ω 2...
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lec9-2 - Linear second-order differential equations with...

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