Ph12a_s8_2009

Ph12a_s8_2009 - Ph12a Solution Set 8 November 24, 2009 6.14...

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Unformatted text preview: Ph12a Solution Set 8 November 24, 2009 6.14 The dispersion relation for a system of coupled pendula is given by eq. 90 of section 2.4: 2 ( k ) = g l + 4 K M sin 2 ka 2 The lower cutoff frequency is p g l corresponding to k = 0 while the upper cutoff frequency is given by q g l + 4 K M corresponding to k = a (see figure p.86 of Crawford). We can now evaluate the group velocity: v g = d dk = 1 2 d 2 dk v g = 1 2 8 K M sin ka 2 cos ka 2 plugging in k = 0 for the lower cutoff frequency we find the that sin ka 2 factor is zero, and thus v g = 0. Taking k = a the cos ka 2 term becomes zero and so the group velosity is zero at the high frequency cutoff as well. The phase velocity at the two frequencies can be found simply from: v = k at k = 0 this yields an infinite phase velocity. At k = a we have v = a q g l + 4 K M A nice graph of the dispersion relation in question is on page 86 of the text. One can read off the group velocity at a particular frequency by looking the slope of the tangent to the graph at the point ( k, ). The phase velocity can be read off of the diagram by calculating the slope of the line connecting the point representing the frequency ( k, ) with the origin. 1 Ph12a Solutions Problem Set 8 November 24, 2009 6.18 (a) The index of refraction is given by: n = c/v = ck/ . So: n 2 = c 2 k 2 2 = 1 + 2 p 2 o- 2 . n 2 is an increasing function of on 0 < o and o < < ....
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Ph12a_s8_2009 - Ph12a Solution Set 8 November 24, 2009 6.14...

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