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Unformatted text preview: 1.7 2D oscillations Consider the motion of a particle with two degrees of freedom. Take the restoring force to be proportional to the distance of the particle from a force center located at the origin, and to be directed toward the origin: F = k r (113) which can be written in polar coordinates: F x = kr cos = kxF y = kr sin = ky (114) with equations of motion x + 2 x = 0 (115) y + 2 y = 0 (116) As before, = k/m . The solutions are x ( t ) = A cos( t ) (117) y ( t ) = B cos( t ) (118) The motion is one of SHO in each of the two directions: both having the same frequency but possibly two different amplitudes and phases. We can obtain the equation for the path y ( t ) = B cos[ t + (  )] (119) = B cos( t ) cos(  ) B sin( t ) sin(  ) (120) Define  at note that cos( t ) = x/A , y = B A x cos  B 1 x 2 A 2 sin (121) or Ay Bx cos = B A 2 x 2 sin (122) squaring gives A 2 y 2 2 ABxy cos + B 2 x 2 cos 2 = A 2 B 2 sin 2  B 2 x 2 sin 2 (123) and B 2 x 2 2 ABxy cos + A 2 y 2 = A 2 B 2 sin 2 (124) If we set = / 2 this gives the equation of an ellipse x 2 A 2 + y 2 B 2 = 1 , = / 2 (125) where in the special case of A = B we have circular motion. Another special case occurs for = 0, B 2 x 2 2 ABxy + A 2 y 2 = 0 , = 0 (126) 14 Factoring ( Bx Ay ) 2 = 0 (127) a straight line, y = xB/A . In the more general case of twodimensional oscillations, the angular frequencies for the motion in the x and y directions are not equal, so that x ( t ) = A cos( x t ) (128) y ( t ) = B cos( y t ) (129) The paths of motion are called...
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 Spring '10
 RogerMelko

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