410ahw7soln - HW#7Solutions Phys410Fall 2011

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HW#7—Solutions —Phys410—Fall 2011 Prof. Ted Jacobson Room 4115, (301)405-6020 www.physics.umd.edu/grt/taj/410a/ [email protected] 11.32 (CO 2 vibrations ) (a) L = 1 2 m ( ˙ x 2 1 + ˙ x 2 3 ) + 1 2 M ˙ x 2 2 - 1 2 k [( x 1 - x 2 ) 2 + ( x 3 - x 2 ) 2 ]. m ¨ x 1 = - k ( x 1 - x 2 ), m ¨ x 2 = k ( x 1 + x 3 - 2 x 2 ), m ¨ x 3 = - k ( x 3 - x 2 ). These take the matrix form M ¨ x = - Kx , with x = ( x 1 ,x 2 ,x 3 ) T , and M = m 0 0 0 M 0 0 0 m , K = k 1 - 1 0 - 1 2 - 1 0 - 1 1 . Now let’s adopt units with k = 1 and m = 1. Then K - ω 2 M = 1 - ω 2 - 1 0 - 1 2 - ω 2 M - 1 0 - 1 1 - ω 2 and det( K - ω 2 M ) = ω 2 (1 - ω 2 )( 2 - (2 + M )), so the normal mode frequencies are ω = 0 , 1 , p 1 + 2 /M . ω = 0 we have x 1 = x 2 = x 3 . This is the translational mode. For ω = 1 we have x 2 = 0 and x 3 = - x 1 . This is the asymmetric mode. Note the center of mass remains at rest. These two are easily guessed. For ω = p 1 + 2 /M , we can insert the value of ω 2 into K - ω 2 M and find the zero eigenvector. This yields x 1 = x 3 and x 2 = - (2 /M ) x 1 , so the two ends move identically and the middle one moves oppositely with relative amplitude 2 a . This has the highest frequency, because for a given displacement of one of the end atoms, the spring is deformed more because of the opposite motion of the middle atom, so the effective spring constant for that atom is larger. (d) For CO 2 , M = M/m = 3 / 4, so the frequency ratio is p 1 + 2 /M = p 1 + 8 / 3 = p 11 / 3 = 1 . 91. The measured frequency of the asymmetric mode (oxygens move op- posite) is listed as 2349 in units of cm - 1 , while that of the symmetric mode (oxygens move in same direction) is 1388. The ratio is 1.69, which is smaller than the above result by 12%. [The website writes 1288 in the text and 1388 in the diagram. Other
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This note was uploaded on 12/29/2011 for the course PHYSICS 410 taught by Professor Jacobson during the Fall '11 term at Maryland.

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410ahw7soln - HW#7Solutions Phys410Fall 2011

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