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#55
Jun1005, 12:59 PM

Mentor
P: 41,438

I can't follow your arithmetic. Start by computing [itex]\cos \omega t \sin \omega t[/itex] for the two times that you have. I recommend you express the times as I did in post #51.



#56
Jun1005, 01:13 PM

P: 171

okay my times expressed the way you suggested are:
0.212 cos(3.70)(0.212)=0.9999 sin(3.70)(0.212)=0.13689 thus: coswtsinwt cos(0.9999)sin(0.13689) =2.3888e3 1.06 cos(3.70)(1.06)=0.9976 sin(3.70)(1.06)0.06839 thus: coswtsinwt cos(0.9976)sin(0.06839) =1.1934e3 now as i mentioned time and time again...here is where i am stuck. i do not know which time or time to plug into my full d(KE)/dt expression[B] 


#57
Jun1005, 01:23 PM

Mentor
P: 41,438

Why not plug in both and see which is bigger? That's what finding the maximum means. 


#58
Jun1005, 01:42 PM

P: 171

okay my times expressed the way you suggested are:
0.212 cos(3.70)(0.212)=0.707812236 sin(3.70)(0.212)=0.70640062 coswtsinwt=0.4999999002 1.06 cos(3.70)(1.06)=0.710627 sin(3.70)(1.06)=0.703568948 coswtsinwt=0.499975091 so now that i manipulate the d(KE)/dt i get: KA^2wcoswtsinwt A=0.0455 m K=3.76 N/m 1.06: 3.891876105e3 J/s 0.212: 3.892069922e3 J/s so if i did this correctly both times seem to be similar so either i did it wrong or it doesnt matter which time you choose 


#59
Jun1005, 01:59 PM

Mentor
P: 41,438




#60
Jun1005, 02:07 PM

P: 171

However when i put that number in :
3.89e3 J/s it tells me the answer is incorrect...ne other suggestions?? 


#62
Jun1005, 02:26 PM

P: 171

I have twice and i still get the same answer....where are you looking... or can you point it out because this questions has been driving me up the wall for almost two days. So if possible can you point out exactly where i went wrong because obviously you know that i know what i am doing.



#64
Jun1005, 02:35 PM

P: 171

I dont believe i am...i am using the exact d(KE)/dt equation that i posted a few posts ago. So as far as i know i am not. What numeric value do you get.? If you choose to calculate it i will redo the question showing you exactly what i am doing.



#65
Jun1005, 02:38 PM

P: 171

1.06
cos(3.70)(1.06)=0.710627 sin(3.70)(1.06)=0.703568948 coswtsinwt=0.499975091 so now that i manipulate the d(KE)/dt i get: KA^2wcoswtsinwt A=0.0455 m K=3.76 N/m (3.76)(2.07025e3)(3.70)(0.710627)(0.703568948) =0.0143999 J/s 


#66
Jun1005, 02:39 PM

P: 171

You are a good man DOC AI...a very good man
thank you so much for your help...i probably drove you up the wall with this question but you stood by me truly a super mentor 


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