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Unformatted text preview: (1 , 0) and of the radius 3 . 5 1 = 2 . 5, and comes to the point (1 2 . 5 , 0) = ( 1 . 5 , 0). From this point, the mass goes in the upper half plane from left to right along the semicircle centered at ( 1 , 0) and of the radius 1 ( 1 . 5) = 0 . 5, and comes to the point ( 1 + 0 . 5 , 0) = ( . 5 , 0). Here it comes to rest because  . 5  &lt; 1, and there is not a lower semicircle starting at this point. See the colored curve in Figure B.42 of the text. EXERCISES 5.5: Coupled MassSpring Systems, page 284 1. For the mass m 1 there is only one force acting on it; that is the force due to the spring with constant k 1 . This equals k 1 ( x y ). Hence, we get m 1 x = k 1 ( x y ) . 307...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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