19()( )222112TMX mXrXrθ≈+ ++±±±( )111222mXmr Xr MXr mrθθ+ ≈ ++±±±±and the M-matrix is Mmmm≈MThe potential energy is 2001122gVkrMmrθω≈+=+±2ΩThe K-matrix is thus 202000Mmω≈ΩKTo solve for 2, we set 20−=KM20200Mmωω−−=−Ω−which yields … 442000mMmΩ+ − Ω−−=422222000MM+Ω+Ω =Neglecting m with respect to M and simplifying yields … 42222200 000−+Ω+Ω=Solving for 2… 220024+Ω−Ω=±222220ΩThus, we get … 2102and==Now, we solve for the amplitudes aof the normal mode vectors … 20a2012200aa=−Using the first of the above matrix equations for 120gives … 2101aandaUsing the second equation for
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.