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Unformatted text preview: Physics 325 Fall 2006: Solutions to Homework Assignment #6 1)Determining x ( t ) and y ( t ) The only forces acting on the mass are from the springs. Using Newtons law F x = kx kx = m ¨ x m ¨ x = 2 kx Similarly m ¨ y = 2 ky . The general solutions for these two differential equations are x ( t ) = A x cos( ωt ) + B x sin( ωt ) y ( t ) = A y cos( ωt ) + B y sin( ωt ) where ω = q 2 k m Plug in the initial conditions: x (0) = A x = x ˙ x (0) = ω · B x = V x to give A x = x andB x = v x ω and A y = y andB y = v y ω Then the solutions are x ( t ) = x cos( ωt ) + v x ω sin( ωt ) y ( t ) = y cos( ωt ) + v y ω sin( ωt ) b)Setting v x = 0, we have x ( t ) = x cos( ωt ) y ( t ) = y cos( ωt ) + v y ω sin( ωt ) To make the trajectory a circle, we require x 2 ( t ) + y 2 ( t ) = R 2 , which means y = 0 and x = ± v y ω in this case. So the conditions for the circular trajectory is x = ± v y ω and y = 0. Therefore x ( t ) = x cos( ωt ), y ( t ) = ± x sin( ωt ). c) To make the trajectory a line, we require y ( t ) x ( t ) = a = const, which means v y in this case. So the condition for the linear trajectory isthis case....
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This note was uploaded on 02/23/2009 for the course PHYS 325 taught by Professor Lamb during the Fall '09 term at University of Illinois at Urbana–Champaign.
 Fall '09
 Lamb
 mechanics, Force, Mass, Work

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