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Unformatted text preview: CHAPTER 2 1. We have ( x ) = dkA ( k ) e ikx = dk N k 2 + 2 e ikx = dk N k 2 + 2 cos kx because only the even part of e ikx = cos kx + i sin kx contributes to the integral. The integral can be looked up. It yields ( x ) = N e  x  so that  ( x )  2 = N 2 2 2 e 2  x  If we look at  A ( k ) 2 we see that this function drops to 1/4 of its peak value at k = .. We may therefore estimate the width to be k = 2 . The square of the wave function drops to about 1/3 of its value when x = 1/2 . This choice then gives us k x = 1. Somewhat different choices will give slightly different numbers, but in all cases the product of the widths is independent of . 2. the definition of the group velocity is v g = d dk = 2 d 2 d (1/ ) = d d (1/ ) =  2 d d The relation between wavelength and frequency may be rewritten in the form 2 2 = c 2 2 so that 2 d d = c 2 = c 1 ( / ) 2 3. We may use the formula for v g derived above for = 2 T  3/2 to calculate v g =  2 d d = 3 2 2 T 4. For deep gravity waves, = g / 2  1/2 from which we get, in exactly the same way v g = 1 2 g 2 . 5. With = k 2 /2 m , = / m and with the original width of the packet w(0) = 2 , we have w ( t ) w (0) = 1 + 2 t 2 2 2 = 1 + h 2 t 2 2 m 2 2 = 1 + 2 h 2 t 2 m 2 w 4 (0) (a) With t = 1 s, m = 0.9 x 1030 kg and w (0) = 106 m, the calculation yields w (1) = 1.7 x (1) = 1....
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 Spring '05
 mokhtari

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