Phys 325 Spring 2011 Lecture 15 - Physics 325 Lecture 15...

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Physics 325 Lecture 15 Impulse Response In the previous lecture, we investigated the response of an oscillator to a periodic driving force. Not all forces are periodic. One can have a force that is impulsive. An impulse is just a force that lasts for a short time and doesn’t repeat. The quintessential case is a force that looks like this: Recall from Physics 111/211 that an impulse causes a change in momentum p f p f t t Therefore, for an oscillator originally at rest, the response to an impulse like this is equivalent to a free oscillator with an initial velocity of 0 f t v m We will solve this problem for such an impulse. Once we’ve accomplished that, we can solve a more general problem of a non-periodic driving force because we can represent it as a sum of impulses and use the principle of superposition. Here we go. Consider an underdamped harmonic oscillator that at t 0 is at rest at 0 x . At t t , it is subjected to a sharp impulse: 0 t 0 t ( ) f t impulse ( ') F t t
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Immediately after the impulse, the oscillator has a velocity of ( ) ( ) ( ) t t t t t t F t F t t v a t dt dt m m     (15.1) For small t , ( lim 0 t   ), we have ( ) ( 0) 0 x t t x t  and the underdamped solution with the initial velocity given by Equation (15.1) is ( ) 1 ( ) ( ) sin ( ) t t d d F t t x t e t t m   (15.2) where d is defined in Equation (12.7). Now suppose that we have a series of impulses of duration t at time t i . Then the response is the superposition of solutions like (15.2): ) 1 ( ) 1 ( ) sin ( ) i N t t i d i i d F t t x t e t t m   (15.3) In the limit where t 0 , the sum in Equation (15.3) becomes an integral ') 0 sin ( ') 1 ( ) ( ') ' t t t d d e t t x t F t dt m   (15.4)
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