ch11 - C 1 I’ILOMCcJOrKJ smﬁm 2 f g constant is the...

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Unformatted text preview: C ' 1/ I’ILOMCcJOrKJ smﬁm 2; f g constant is the ratio of applied to t. k=5=.lee7sn . - ~ x .0.85m—,0.65m‘ 0.20m ~ Thespting constant is found from the ratio of appliedvfoi'oeyto‘disﬁlécemént.‘ .=£=2€.= (68 “)(9'fm/s = 1.3335105‘N/m f ‘3' ""‘=-*- “41333520: m»=1'-467st* 1 , 2:: m 2:: 9. (a) 7 At equilibﬁum, they veldcity is its maximum. ,v...-\/;A—wA-2”fA=21t(3Hz)(0.13m);V364§9m/s§ (b) _ me Equation. (11-5);we ﬁndthe velocity at anxpasiﬁon. » - 2 ' - 2 . r ‘ V " "i ' - A: h _) ’(0.13m)2 4156M,” I . : , (c) 5» #va =+(o.eo kg)(-2.4sm/s)z'=1.sou.‘ ' " . I _ Q). hasramaximumdispl'acem'em‘att; o,th¢pgsiﬁ9n I .bgdby the 5 (0-13,9)'M6.%t) . 3 s _ x=(0.g3m)om(2z(3ﬂHz)r).+ x=- '16. msgeneralformofthemoﬁonié gm“ =0.450036.40:._ '(‘l'ly'The'amnplimde is A =V.Ic__’II = . (b) 'mﬁequmcyisfoundby’a)=2af‘=6.4os" .9 f-_-. - ‘l ' “ .. 1'. _ 640‘s '=1.019Hz~1’.0'21§z 2:: V (c) MWencrgyisgivenby , r .. 7 r U Eu- ‘ém‘i— =%"‘(."’A)’ =%(0.60ks)[(6-4° 8")(0-45 m)]’ = 24881 ~ - (d) Mpotendalmgyisgim-by 2 ~- ._ EN :99 =-2mezxz =-;-(0.60 kg),(’6t40,sf')2 (030 m)? 1,111; L ......... -w .._' 22. (a) ForAJheamplimdeisAAém.‘ (b) ForA,meﬁequency is 1_cy91c_ 4'0 seconds, fA =- cycle’eVWZLQVW-S? ‘ 3 (c) For C,theperi'odis .TA =.. ForB,'mep¢1iodié’T, = M (d) Object A has a dispmmem‘of 0 men; t -‘-='"Q;,7'861_i_tui_s g IA gammy“) —) v y when 1:07-7:30. itis aoosine function. ...
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ch11 - C 1 I’ILOMCcJOrKJ smﬁm 2 f g constant is the...

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