3.6112.59.826lTsgππ==≈Ds3.7For springs tied in parallel: () ( )1212sFxkxkxk kx=−−+()12kkmω+=For springs tied in series: The upward force is meqkx. Therefore, the downward force on spring is 2keqkx. The upward force on the spring is 2k1kx′where x′is the displacement of P, the point at which the springs are tied. Since the spring is in equilibrium, 2k1eqk′ =x. Meanwhile, The upward force at P is 1′. The downward force at P is ( )2kxx′−. Therefore, kx x′′
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.