## SHM Equation problem

okay but are any of my answers correct, and which method would i have to employ to get the correct answer?

Mentor
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 Quote by DDS the times which i plugged in and are correct are as follows: 0.212s and 1.06 s this is what i have done thus far: A=0.0455 m K=3.76 N/m w=3.70 (3.76)*(0.0455)^2*(3.70)cos(3.70)(0.212)*sin(3.70)(1.06) 7.78414e-3[3.699653267][0.068398368] =1.953e-3
To get a meaningful answer, use a single value for t. (For some reason, you used both values for t in the same expression.)
 (3.76)*(0.0455)^2*(3.70)cos(3.70)(0.212)*sin(3.70)(0.212) 7.78414e-3[3.699653267][0.013689935] =3.94e-4 J/s which is wrong and (3.76)*(0.0455)^2*(3.70)cos(3.70)(1.60)*sin(3.70)(1.60) 7.78414e-3[3.680267438][0.068398368] =1.959e-3 which is also wrong this is what ive been trying to get acros no matter how i try to plug it in i get it wrong, ive even tried your way of simplifing the time experison
 Mentor Blog Entries: 1 I can't follow your arithmetic. Start by computing $\cos \omega t \sin \omega t$ for the two times that you have. I recommend you express the times as I did in post #51.
 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) =2.3888e-3 1.06 cos(3.70)(1.06)=0.9976 sin(3.70)(1.06)0.06839 thus: coswtsinwt cos(0.9976)sin(0.06839) =1.1934e-3 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]

Mentor
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 Quote by DDS 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
Incorrect. Realize that $\omega$ = 3.7 radians/sec, not degrees/sec.
 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]
No point in worrying about that until you get your calculations correct.

Why not plug in both and see which is bigger? That's what finding the maximum means.
 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: 3.891876105e-3 J/s 0.212: 3.892069922e-3 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

Mentor
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 Quote by DDS 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
Right! The value of d(KE)/dt is exactly the same for both times.
 However when i put that number in : 3.89e-3 J/s it tells me the answer is incorrect...ne other suggestions??
 Mentor Blog Entries: 1 Recheck your arithmetic.
 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.
 Mentor Blog Entries: 1 I suspect you are dropping an $\omega$.
 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.
 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)(2.07025e-3)(3.70)(-0.710627)(-0.703568948) =0.0143999 J/s
 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