p2spring07_SOLUTIONS_ONLY - some answers: 1a) 2 cycles per...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
some answers: 1a) 2 cycles per second, so ω = 4 π 1b) m = 100 000, k = 80 000 g/. 02 = 4(10) 6 g .02 80 000g ky y 1c) Since y = . 003 sin( ωt ), we get my ±± + ky = . 003( - ω 2 m + k ) sin( ωt ), so f 0 = . 003( - ω 2 m + k ). (It was not required but this works out to . 003( - 16 π 2 10 5 + 4(10) 6 g ) . = . 003( - 16 + 40)(10) 6 . = 70 000 Newtons approx.) 1d) The empty car has unforced equation 20 000 y ±± + ky = 0 with circular frequency q k 20 000 = q 80 000 g ( . 02)20 000 , while the pendulum has equation θ ±± = - g l θ with circular frequency p g l . These agree when 1 l = 4 . 02 , or l = . 005 meter. 2) The characteristic polynomial is 2 r 2 + 5 r + 3 = (2 r + 3)( r + 1), so the general solution to the homogeneous equation is y c = c 1 e - x + c 2 e - 3 2 x . Un- determined coefficient method gives a particular solution y p = - 4 25 sin(2 x ) - 8 25 cos(2 x ) - 2. Adding these, then applying initial conditions we find y ( x ) = 15 25 e - x + 18 25 e - 3 2 x + y p ( x ). 3) Integrate four times (definite integrals save space and time) to find EIy ( x ) = Lx 5 120 - x 6 360 - L 3 x 3 36 + L 4 x 2 24 and at the end EIy ( L ) = L 6 ( 1 120 - 1 360 - 1 36 + 1 24 ) . Since 120 < 360 and 24 < 36, the number in parenthesis is positive.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/21/2011 for the course MATH 2930 taught by Professor Terrell,r during the Spring '07 term at Cornell University (Engineering School).

Ask a homework question - tutors are online