t 2 m 500 628 126 s k 125 789242382

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Unformatted text preview: in internal energy over cycle = −(0 − 233) = 233kJ WC →A = 233kJ ￿ Alternatively could use W = − P dV and integrate the expression. This takes longer. ! ! "! ! PHYS1131 Exam 2012-T2 Oscillation question [25] marks (a) (i) At the moment of release:! " F = #kx = ma 125 $ 0.687 = 5.0a 85.9 a= = 17.17 = 17.2m / s2 5.0 !!!!!!!!#$$%!&'(!)(*$+,!+-!+./$001&$+2!-+*!&'(!.)*$23!.4.&(56! ! ! !!!!!!!!!!!!!!!! T = 2 " #$$$% m 5.00 = 6.28 # = 1.26 s !!!!!!!!!!!!! k 125 7'(!8+*9!,+2(!:4!&'(!.)*$23!:(&8((2!;!<!=>?@A!12,!=6! ! !!!!!!!!!!!!!! W = F .x = 0.687 " 0 1 kxdx = k x 2 = 29.5 J ! 2 #$B%!!!!! ! 1 = 0.794 Hz T !!!!!!!!!!!!!!! 2T = 2.52 s f= ! displacement x = 0.687cos(2 "ft) !!!!!! !! ! ! ! ! !!!!!!C5100!,15)$23!!#:!<!=>DE!F!2+&!-+*!51*9$23%! ! !!! ! ! ! ! ! ! ! ! ! ! ! ! ! ! G! ! #:% !#$%!H-!&'(!)1&'!,$--(*(2/(!*(.I0&.!$2!&8+!)(*$+,.!1**$B10!,$--(*(2/(!-+*!&'(! 81B(!&*1$2.!-*+5!(1/'!.+I*/(J!$&!5I.&am...
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This note was uploaded on 05/31/2013 for the course PHYSIC 1131 taught by Professor John during the Three '13 term at University of New South Wales.

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