Solve SHM Eqn Problem: Max Energy Transformation

[Doc Al]

If you express your times like this: \pi/(4 \omega) and (5 \pi)/(4 \omega), you will have an easier time computing the values of d(KE)/dt. The units will be energy/time.

 

[DDS]

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

 

[Doc Al]

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.)

 

[DDS]

(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 I've been trying to get acros no matter how i try to plug it in i get it wrong, I've even tried your way of simplifing the time experison

 

[Doc Al]

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.

 

[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

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

 

[Doc Al]

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

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]

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 doesn't matter which time you choose

 

[Doc Al]

so if i did this correctly both times seem to be similar so either i did it wrong or it doesn't matter which time you choose

Right! The value of d(KE)/dt is exactly the same for both times.

 

[DDS]

However when i put that number in :

3.89e-3 J/s it tells me the answer is incorrect...ne other suggestions??

 

[Doc Al]

Recheck your arithmetic.

 

[DDS]

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]

I suspect you are dropping an \omega.

 

[DDS]

I don't 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]

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]

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