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Unformatted text preview: PHYS 2214  Fall 2008 PS2 Solutions PHYS 2214: Waves, Optics, and Particles Fall 2008 Problem Set 2 Solutions September 9, 2008 c Joe P. Chen 1. (a) To find the extremum of amplitude A = A ( ) we need to set ( dA/d ) = 0. Since A =  A  = F m 1 ( 2 2 ) + i (2 / ) = F m 1 q ( 2 2 ) 2 + (2 / ) 2 (1) we have 0 = dA d = d d F m 1 q ( 2 2 ) 2 + (2 / ) 2 = F m 2 2 2 + ( 2 / 2 ) h ( 2 2 ) 2 + (2 / ) 2 i 3 / 2 (2) The realvalued solution to Eq. (2) is = 0 or 2 2 + (2 / 2 ) = 0. But as you may recall from the resonance curve drawn in lecture, = 0 corresponds to a local minimum of A ( ), whereas the latter solution gives , where A ( ) is approximately maximal. Therefore the resonant frequency of a driven damped oscillator with natural frequency and amplitude decay time is r = r 2 2 2 which is less than as advertised 1 . (b) For light damping ( 1 or 1 / ( ) 1), we use the approximation that for a dimensionless number x 1, (1 + x ) n 1 + nx (3) Applying this to r we get r = s 1 2 ( ) 2 1 1 2 2 ( ) 2 = 1 1 ( ) 2 which is valid to second order in the small parameter 1 / ( ). (c) To plot A as a function of / , we make some tweaks to Eq. (1) and rewrite it as A = F m 1 q ( 2 2 ) 2 + (2 / ) 2 = F m 2 1 s 1 2 2 + h 2 1 i 2 Notice that the square root in the denominator is now dimensionless, so the prefactor F / ( m 2 ) must carry the same dimension as A , namely, length. (Check it explicitly!) We can now make what is called a figure of merit where, for a given quality factor (here we set Q = 1 2 = 4), we plot A in units of F / ( m 2 ) as a function of / , so that the values on both axes of the plot become strictly dimensionless (Fig. 1(a)). This is an extremely useful way of collapsing all resonance curves with different values of the physical variables F , m , and onto one single curve. From the plot we see that the amplitude A is maximal when / 1. 1 What if < 2 / ? How do you interpret this scenario physically?...
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 Fall '07
 GIAMBATTISTA,A

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