SHM Equation problem


by DDS
Tags: equation
Doc Al
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#55
Jun10-05, 12:59 PM
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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.
DDS
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#56
Jun10-05, 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)
=2.3888e-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]
Doc Al
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#57
Jun10-05, 01:23 PM
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Quote 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 [itex]\omega[/itex] = 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.
DDS
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#58
Jun10-05, 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:
3.891876105e-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
Doc Al
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#59
Jun10-05, 01:59 PM
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Quote 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.
DDS
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#60
Jun10-05, 02:07 PM
P: 171
However when i put that number in :

3.89e-3 J/s it tells me the answer is incorrect...ne other suggestions??
Doc Al
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#61
Jun10-05, 02:18 PM
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Recheck your arithmetic.
DDS
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#62
Jun10-05, 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.
Doc Al
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#63
Jun10-05, 02:32 PM
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I suspect you are dropping an [itex]\omega[/itex].
DDS
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#64
Jun10-05, 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.
DDS
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#65
Jun10-05, 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)(2.07025e-3)(3.70)(-0.710627)(-0.703568948)
=0.0143999 J/s
DDS
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#66
Jun10-05, 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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