This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 θ = kxF y = kr sin θ = ky (114) with equations of motion ¨ x + ω 2 x = 0 (115) ¨ y + ω 2 y = 0 (116) As before, ω = k/m . The solutions are x ( t ) = A cos( ω t α ) (117) y ( t ) = B cos( ω t β ) (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[ ω t α + ( α β )] (119) = B cos( ω t α ) cos( α β ) B sin( ω t α ) sin( α β ) (120) Define δ ≡ α β at note that cos( ω t α ) = x/A , y = B A x cos δ B 1 x 2 A 2 sin δ (121) or Ay Bx cos δ = B √ A 2 x 2 sin δ (122) squaring gives A 2 y 2 2 ABxy cos δ + B 2 x 2 cos 2 δ = A 2 B 2 sin 2 δ B 2 x 2 sin 2 δ (123) and B 2 x 2 2 ABxy cos δ + A 2 y 2 = A 2 B 2 sin 2 δ (124) If we set δ = ± π/ 2 this gives the equation of an ellipse x 2 A 2 + y 2 B 2 = 1 , δ = ± π/ 2 (125) where in the special case of A = B we have circular motion. Another special case occurs for δ = 0, B 2 x 2 2 ABxy + A 2 y 2 = 0 , δ = 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( ω x t α ) (128) y ( t ) = B cos( ω y t β ) (129) The paths of motion are called...
View
Full
Document
This note was uploaded on 09/20/2010 for the course AMATH 261 taught by Professor Rogermelko during the Spring '10 term at Waterloo.
 Spring '10
 RogerMelko

Click to edit the document details