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Physics 325 Lecture 10

# Physics 325 Lecture 10 - Physics 325 Lecture 13 Driven...

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Physics 325 Lecture 13 Driven Oscillations Now we move to the case of oscillators with driving forces. We will start with sinusoidal driving forces and linear velocity-dependent damping. The equation of motion is 0 cos mx bx kx F t ω + += ±± ± (13.1) Here k is the spring constant and is the frequency of the driving force, which can be different from the resonant or natural frequency, 0 km = , of the oscillator. Dividing Equation (13.1) through by m and defining β= b/2m as before, and A=F 0 /m, we get 2 0 2c o s x xx A t β ++= ± (13.2) The procedure for finding the general solution to this differential equation is to first find the “complementary” solution, which is the solution to the corresponding homogenous equation with the right hand side set to zero, and then find the particular solution which actually solves the non-homogenous equation. This may sound strange, but since it is a linear equation, we can add solutions. Adding the solution to the homogenous equation does not affect the particular solution yielding the answer we want. We have already solved the complementary equation. That’s just our damped oscillator without a driving force. So the complementary function is () 22 0 12 t t c xt e A e A e βω −− 0 t =+ (13.3) which is Equation (12.9). A reasonable guess for the particular solution is ( ) cos sin p x tB t (13.4) which has the same frequency as the driving frequency. This makes sense since the complementary solution dies out at large times since it corresponds to a damped oscillator. We substitute Equation (13.4) into Equation (13.2) to get 1 2 01 02 cos sin 2 sin 2 cos cos sin cos B t B t B t BtBt At ωω −−− + ++ = The terms on the left that have sin t must sum to zero, and the terms that have cos t must sum to A : 0 1 21 0 2 2 20 B BB B ωβ −+ + = A

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And finally () 22 0 12 00 2 A A BB ωω βω ω ωβω ωβ == −+ So the particular solution is not exactly simple: ( ) 0 2 2 0 cos 2 sin 2 p At A xt t β = (13.5) To understand this solution better, we would like to be able to write it as some sinusoidal piece with a phase and an amplitude: ( ) ( ) cos p xt D t δ = (13.6) Using our trig identities we expand this as cos cos sin sin p t D t δω =+ Comparing with Equation (13.5), we can write 0 2 2 0 2 2 0 cos 2 2 sin 2
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Physics 325 Lecture 10 - Physics 325 Lecture 13 Driven...

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