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Unformatted text preview: Physics 325 Lecture 16 Impulse Response (cont.) What do we do when v ? Lets consider an undamped case for which we need to solve 2 ( ) x x f t (16.1) with initial conditions (0) 0, (0) x x v We cannot write the solution as ( ) t x t f t G t t dt (16.2) because this gives x t for all t when ( ) f t , whereas we have x t v . The solution is to add to the integral (16.2) the solution to the homogenous equation ( ) sin v x t t This is similar to our technique for the driven harmonic oscillator where the transient solution was needed to satisfy the initial conditions though here there is nothing transient about the homogenous solution. The complete solution, assuming that the driving acceleration f ( t ) becomes nonzero for some t , is ( ) sin t v x t t f t G t t dt (16.3) Obviously this works for ( ) f t , lets check it in the case of an impulsive force. f t ( ) f t t f t ( ) f t t 1 ( ) sin f sin ( ') ' t t t v x t t t t dt Then for t t : ( ) sin sin ( ) v f t x t t t t t (16.4) Lets check the result. At t t , the mass is at 0 0 ( ) sin v x t t with velocity 0 0 ( ) cos x t v t The impulse is applied from t t to t t t , and at t t t the mass is at 0 0 ( ) sin sin v x t t t t v t with velocity 0 0 ( ) cos cos cos x t v t t f t t v t f t But f t F t m v , so after the impulse the oscillator is moving at x t v as expected. The subsequent motion of the oscillator is just the motion that we have already solved for but with a new set of initial conditions 0 0 sin v x t t t x t t v v Starting the clock at t t by defining t t t , we can then write the equations of motion as 0 0 0 0 cos sin cos sin v f t v t x t t t t We can rearrange this as 0 0 0 0 sin cos cos sin sin sin sin sin sin v f t x t t t t t t v f t t t t v f t t t t...
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This note was uploaded on 10/06/2011 for the course PHYS 325 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Staff
 Physics, mechanics

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