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Phys 325 Spring 2011 Lecture 16

# Phys 325 Spring 2011 Lecture 16 - Physics 325 Lecture 16...

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Physics 325 Lecture 16 Impulse Response (cont.) What do we do when 0 0 v ? Let’s consider an undamped case for which we need to solve 2 0 ( ) x x f t (16.1) with initial conditions 0 (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   0 x t for all t when ( ) 0 f t , whereas we have 0 0 x t v . The solution is to add to the integral (16.2) the solution to the homogenous equation 0 0 0 ( ) 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 non-zero for some 0 t , is 0 0 0 0 ( ) sin t v x t t f t G t t dt (16.3) Obviously this works for ( ) 0 f t , let’s check it in the case of an impulsi ve force. 0 f t ( ) f t 0 t 0  

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0 0 0 0 0 0 0 0 1 ( ) sin f sin ( ') ' t t t v x t t t t dt  Then for 0 t t : 0 0 0 0 0 0 0 0 ( ) sin sin ( ) v f t x t t t t t (16.4) Let’s check the result. At t t 0 , the mass is at 0 0 0 0 0 ( ) sin v x t t with velocity 0 0 0 0 ( ) cos x t v t The impulse is applied from t t 0 to 0 t t t   , and at 0 t t t   the mass is at 0 0 0 0 0 0 0 0 0 ( ) sin sin v x t t t t v t     with velocity 0 0 0 0 0 0 0 0 0 0 ( ) cos cos cos x t v t t f t t v t f t   But 0 0 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 0 0 0 0 0 sin v x t t t x t t v v   Starting the clock at 0 t t by defining 0 t t t , we can then write the equations of motion as   0 0 0 0 0 0 0 0 0 0 0 cos sin cos sin v f t v t x t t t t  
We can re-arrange this as   0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 which is the same as the Green’s function solution in Equation (16.4).

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Phys 325 Spring 2011 Lecture 16 - Physics 325 Lecture 16...

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