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Unformatted text preview: 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 cos mx bx kx F t (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, k m , of the oscillator. Dividing Equation (13.1) through by m and defining b 2 m as before, and A F m , we get 2 2 cos x x x A t (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 that 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. Thats just our damped oscillator without a driving force. So the complementary function is 2 2 2 2 1 2 ( ) t t t c x t e Ae A e (13.3) which is Equation (11.9). A reasonable guess for the particular solution is 1 2 cos sin p x t B t B t (13.4) which has the same frequency as the driving frequency. This makes sense since the complementary solution dies out at large t since it corresponds to a damped oscillator. We substitute Equation (13.4) into Equation (13.2) to get 2 2 2 2 1 2 1 2 1 2 cos sin 2 sin 2 cos cos sin cos B t B t B t B t B t B t A t The sin t terms on the left must sum to zero, and the cos t terms must sum to A : B 1 2 2 B 2 2 B 1 A B 2 2 2 B 1 2 B 2 And finally 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 A A B B So the particular solution is not exactly simple: 2 2 2 2 2 2 cos 2 sin 2 p A t A t x t (13.5) To understand this solution better, we would like to write it as some sinusoidal piece with a phase and an amplitude:...
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