Solve SHM Eqn Problem: Max Energy Transformation

[DDS]

wKA^2 1/2wsin2wt

 

[Doc Al]

wKA^2 1/2wsin2wt

Now you have too many. Take the expression given in #20 and plug in the substitution given in #28.

 

[DDS]

KA^2 1/2wsin2wt

 

[Doc Al]

OK. Keep going.

 

[DDS]

so now if i take the derivative of this i get:

K2A w^2cost

 

[DDS]

can anyone help me from here?

 

[whozum]

Since Doc is offline I think I'll give you a little nudge. From what I've read your trying to find the maximum of kA^2\omega\cos \omega t \sin \omega t which = \frac{1}{2}kA^2\sin 2\omega t.

Once you find the derivative, you'll want to find the maximum of the function, which is either at the critical points of the efunction (points where the derivative = 0) or the endpoints given by the domain.

The last derivative you have is incorrect, try identifying what is constant and what isnt. Don't forget to apply the chain rule when deriving sin(2\omega t) [/tex]

 

[DDS]

These are the derivatives i get:

wKA^2cos(2wt)

or

tKA^2cos(2wt)

i am leaning toward the first one...AM i right?>?>

 

[OlderDan]

These are the derivatives i get:

wKA^2cos(2wt)

or

tKA^2cos(2wt)

i am leaning toward the first one...AM i right?>?>

I suggest you consolidate your work to this point and bring it together in one place. I also suggest that you follow this outline. You have more trig transformations and identities going on than you need.

PE = \frac{1}{2}kx^2

R = \frac{{dPE}}{{dt}} = \ \ ?

and

\frac{{dR}}{{dt}} = \ \ ? = 0

with all of the above expressed in terms of

x \ \ , \ \ \frac{{dx}}{{dt}} \ \ , \ \ \frac{{d^2 x}}{{dt^2 }}

Use the general form of the given equation for x to write

x = A\sin \omega t

\frac{{dx}}{{dt}} = \ \ ?

\frac{{d^2 x}}{{dt^2 }} = \ \ ?

expressed in terms of A and \omega . Now substirute these into the equation

\frac{{dR}}{{dt}} = \ \ ? = 0

and solve it for t.

 

[DDS]

Excellent , i have gotten the answer to part A but in a different manner.
I determined that :

Period= 2*pi/w

however the maximum shift will occur at t=T/8 and the second time
t'=T/2 + T/8

now I've come to part be and i don't even know what to do.

i am thinking that i must take the second derivative and set it ot zero but i am not sure, any help for part b of this question??

 

[Doc Al]

Now that you've found the times where d(KE)/dt is maximum, just plug those times into your expression for d(KE)/dt to find out what that maximum value is.

 

[DDS]

so i just plug those times into this equation:

KA^2wcoswtsinwt


thus each time taking a spot in each place for t??

 

[DDS]

because when i solved it i did the following:

v(t)=1/2KA^2wsin2t

R(t)=d(V)/dt = -wkA^2 coswtsinwt

dR(t)/dt = -w^2KA^2 (cos^2wt - sin(2wt) =0

coswt=sinwt

cot= pie/4 t=pie/4*w

and then i got my times to be 0.212 and 1.05

 

[DDS]

or do i have to plug in each time sperataly and take the average?

 

[Doc Al]

Have you actually tried plugging in your times to find out what the maximum values are? Do it.

 

[DDS]

I have i tried to plug each time speaqrtly into each variable of t giving me a negative number and if i try pluggin in each time and then taking the average i get a positive answer

so i don't know which way to set it up and thus get the answer

 

[Doc Al]

Are you telling me that the average of two negative numbers is positive?

If you truly have found the proper maxima, the values for d(KE)/dt will be positive.

 

[DDS]

im think we are confusing what i have done:

i have take my two times and either:

pluged both into one equation

and taken one time plugged it into both points in my eqaution did the same for the other time, take the average of the two numbers and that's my energy.

i have two solutions and i don't know which one is correct.

 

[Doc Al]

I'm having a hard time understanding what you are doing. First, you are finding the maximum rate of change of KE, not the maximum energy. You found the values for time that you think correspond to the maximum values of d(KE)/dt. So... plug in your times and see what the rates are.

And don't be vague in your answer:
(1) write the expression you are using for d(KE)/dt.
(2) write the times corresponding to the maximum values of d(KE)/dt.
(3) calculate the maximum values.

 

[DDS]

The expression i am using is:

KA^2wcoswtsinwt


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

or now i take the expression plug in time 0.212 into both places for t and then take my other time 1.06 plug that into both places for t and take and average of the two numbers which is:

1.17095e-3

also what would be the units for the answer?

 

[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??