Solve SHM Eqn Problem: Max Energy Transformation

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The discussion focuses on solving a problem involving a mass on a spring, specifically determining when potential energy (PE) is rapidly transforming into kinetic energy (KE) and finding the maximum rate of energy transformation. Participants emphasize the importance of deriving expressions for both PE and KE as functions of time and using derivatives to find critical points. The correct approach involves maximizing the rate of change of KE, which is linked to the rate of change of PE, rather than directly maximizing PE. After some back-and-forth, a participant successfully identifies the times at which the energy transformation is maximized and seeks guidance on calculating the maximum rate of transformation. The conversation highlights the complexities of applying calculus to harmonic motion problems.
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A mass on a spring with a constant of 3.76 N/m vibrates, with its position given by the equation x = (4.55 cm) cos(3.70t rad/s). During the first cycle, for 0<t<1.70 s, when is the potential energy of the system changing most rapidly into kinetic energy? There are two solutions, enter both with the smaller one first.

B)What is the maximum rate of energy transformation?

I have taken the square of the function given, then i have found the derivate of the sqaured function. I then set the derivate equal to zero in order to find my max and min but i get the wrong answer.

As for b i have no clue where to even start

ANy help would be extremely appriciated
 
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DDS said:
I have taken the square of the function given, then i have found the derivate of the sqaured function. I then set the derivate equal to zero in order to find my max and min but i get the wrong answer.
You need to find where the rate of change of PE is maximum, not where the PE is maximum.
 
How would i go about doing so, this is a very confusing problem at least to me. The only thing i can think of is the slope of the graph, so where the slope being the steepest that would be the greatest rate of change.

How would i go about solving this problem??
 
DDS said:
How would i go about doing so, this is a very confusing problem at least to me. The only thing i can think of is the slope of the graph, so where the slope being the steepest that would be the greatest rate of change.

How would i go about solving this problem??

You know the energy is conserved, so whatever kinetic energy is "lost" is converted to potential and vice versa. If you write the kinetic anergy, and take its derivative with respect to time, that is the rate of change of kinetic energy. It will be a function of time. You can find the maximum of that function in the usual way by finding where its slope (derivative) is zero.
 
DDS said:
How would i go about doing so, this is a very confusing problem at least to me. The only thing i can think of is the slope of the graph, so where the slope being the steepest that would be the greatest rate of change.

How would i go about solving this problem??
You were on the right track. If you write the potential energy (1/2 k x^2) as a function of time (as I think you were trying to do by "squaring" the position function) then you can find its rate of change. (Which is of course it's slope, if you graphed it.) To find where that function has its minimum, take its derivative. (Note: Where the PE decreases fastest is where the KE increases fastest.)

(This is equivalent to OlderDan's suggestion; I just wanted to tie it into what I thought you were doing. Writing the KE as a function of time, as he suggests, may make it easier for you to grasp.)
 
Ihave grouped everyones suggestions and have gotten to this stage:

I have found that

2wt=N(pi)
t= N(pi)/ 2w

i have arbitrarely pluged in vlaues for pie to find times between my given domain, thus i have resulted with these solutions:

n(0)=0
n(1)=0.424
n(2)=0.848
n(3)=1.27
n(4)=1.69

if i plug in 5 for n i get a time that is to large for my interval thus these are the for possible times. However i have tried each combination of these times and they do no produce the right answer.

Can sumone please help me solve this problem?
 
Is there someone who can help me from here?
 
DDS said:
I have found that

2wt=N(pi)
t= N(pi)/ 2w
Show how you obtained this result.
 
y=(Acoswt)^2
y prime= 2 Acoswt + (-sinwt)w =0
through an identity i re wrote it in the form:

2sinwt + coswt=sin2wt=0

sin2wt=0
2wt=N(pi)
t= N(pi)/ 2w

thats the jist of it, now from my equation for t i did what i hsowed a few posts previous,

n(0)=0
n(1)=0.424
n(2)=0.848
n(3)=1.27
n(4)=1.69

i attempted to plug in the values in all possible combinations and i got the answer wrong.

What do i do from here?
 
  • #10
DDS said:
y=(Acoswt)^2
y prime= 2 Acoswt + (-sinwt)w =0
through an identity i re wrote it in the form:

2sinwt + coswt=sin2wt=0
What you seem to be doing (as was pointed out before) is finding the maximum of the PE by (trying to) set its derivative to zero. But that's not the problem. (That's how you find the maximum PE, but that's not the question.)

Start over. Find an expression for the KE as a function of time. (Use conservation of energy.) Then find the rate of change of the KE. That's the function you must maximize.
 
  • #11
So i would have to find the derivative of

1/2mv^2...is there an easier way to solve this problem because i am so discourages now, i did all this work for nothing>?
 
  • #12
DDS said:
...i did all this work for nothing>?
Not for nothing. It is a necessary step, you just stopped too soon.

Hint: KE = Total Energy - PE

Or you can just realize that d(KE)/dt = - d(PE)/dt.

Can you write the PE as a function of t?
 
  • #13
i think you can but i am not sure how to.

I seem to be mentaly fixed on the material i have gone through that i don't know how to manipoulate it to get to where i have to go
 
  • #14
Well... what possessed you to square the function given and take its derivative?

Hint: Spring potential energy is given by 1/2 k x^2, where x is the displacement from equilibrium. You are given x(t). Use it.
 
  • #15
well i squared it due to what you explained :

im know 1/2kx^2 and i am given x so i plug in x into my equation.. and i took the derivate because i know i have to.

but are you saying for me to plug in times bewtween the given interval??

because what i am interpreting from you telling me to use x(t)
that would mean that i would plug in values of t into my position equation.

That doesn't make sense to me seeing as how we are looking for time when the change is the greatest
 
  • #16
DDS said:
well i squared it due to what you explained :

im know 1/2kx^2 and i am given x so i plug in x into my equation.. and i took the derivate because i know i have to.
If you found PE, then took the derivative, you'd have d(PE)/dt. Good! Since d(KE)/dt = -d(PE)/dt, you should have an expression for d(KE)/dt.

Now write that expression as a function of t. (Don't set it equal to zero!)

That's what you need to maximize. (You'll need to take its derivative. Which involves the 2nd derivative of the PE.)
 
  • #17
y=(Acoswt)^2
y prime= 2 Acoswt + (-sinwt)w =0

so as you a said ( am i doing this correctly??)

y prime = - 2 Acoswt + (-sinwt)w is this an expression for d(KE)/dt. ??)

and if it is where do i take it from here
 
  • #18
DDS said:
y=(Acoswt)^2
You want an expression for spring PE:
\mbox{PE} = 1/2 k x^2 = 1/2 k A^2 \cos^2 \omega t
y prime= 2 Acoswt + (-sinwt)w =0
Realize that the derivative of \cos^2 \omega t is -2 \omega \cos \omega t \sin \omega t, not 2 \cos \omega t - \sin \omega t. (I assume you are just writing it wrongly?)

so as you a said ( am i doing this correctly??)
See my correction above.

y prime = - 2 Acoswt + (-sinwt)w is this an expression for d(KE)/dt. ??)
Once you get the correct expression for d(PE)/dt, then yes, -d(PE)/dt will equal d(KE)/dt.

and if it is where do i take it from here
You find the maxima of that function just like any other.
 
Last edited:
  • #19
So is my correct derivative this:

-2coswtsinwt

and thus my term for KE is just: 2coswtsinwt and then i use this equation to find the maxima, and just to calrify, what you mean by maxima is set it equal to zero and find the roots. Because if so I am a littel consfused as to how i would find the maxima of such a function
 
  • #20
As I stated in the last post, the formula for PE is: \mbox{PE} = 1/2 k x^2 = 1/2 k A^2 \cos^2 \omega t

Thus:
d({KE})/dt = -d({PE})/dt = -1/2 k A^2(-2 \omega \cos \omega t \sin \omega t) = k A^2 \omega \cos \omega t \sin \omega t
This is what you must find the maxima of. To find the maxima, treat it as any other function: Set its derivative equal to zero. (Before taking the derivative, simplify using a trig identity.)
 
  • #21
SO just to make sure before i move onto solve this problem:

i must take the dervative of my KE term which is :

KA^2wcoswtsinwt

once i have its derivative , set it equal to zero and solve, once i solve i should get the times at which the change is most rapid. ( and just to clarifiy, we are still working on part A of the question right??)
 
  • #22
Right. (Please simplify that expression first. And call it the rate of change of KE.) And remember, setting the derivative equal to zero finds both maxima and minima, so you have to check. Yep, still on part A.
 
  • #23
would i use the identity of

2sinwt+coswt=sin2wt=o becuase i am not familiar with what identity i would use to simply the problem?
 
  • #24
I assume you meant to write: 2 \sin \omega t \cos \omega t = \sin 2 \omega t. (Watch those + signs!) That's the one to use.
 
  • #25
So is my derivative of the rate of change of Ke this:

-w^2*2A*sinwt*coswt
 
  • #26
How did you get that? Again, first simplify the function using that trig identity, then take the derivative.
 
  • #27
The only thing i can think of simplifying the eqatuion would be to due this

KA^2wcoswtsinwtcoswt

but i don't see how that's simplifying it that much
 
  • #28
Realize that:
\cos \omega t \sin \omega t = \sin \omega t \cos \omega t = (1/2) \sin 2 \omega t.
 
  • #29
so it would just be KA^2 1/2sin2wt ?
 
  • #30
DDS said:
so it would just be KA^2 1/2sin2wt ?
You dropped a few omegas.
 
  • #31
wKA^2 1/2wsin2wt
 
  • #32
DDS said:
wKA^2 1/2wsin2wt
Now you have too many. Take the expression given in #20 and plug in the substitution given in #28.
 
  • #33
KA^2 1/2wsin2wt
 
  • #34
OK. Keep going.
 
  • #35
so now if i take the derivative of this i get:

K2A w^2cost
 
  • #36
can anyone help me from here?
 
  • #37
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]
 
  • #38
These are the derivatives i get:

wKA^2cos(2wt)

or

tKA^2cos(2wt)

i am leaning toward the first one...AM i right?>?>
 
  • #39
DDS said:
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.
 
  • #40
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??
 
  • #41
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.
 
  • #42
so i just plug those times into this equation:

KA^2wcoswtsinwt


thus each time taking a spot in each place for t??
 
  • #43
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
 
  • #44
or do i have to plug in each time sperataly and take the average?
 
  • #45
Have you actually tried plugging in your times to find out what the maximum values are? Do it.
 
  • #46
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
 
  • #47
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.
 
  • #48
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.
 
  • #49
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.
 
  • #50
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?
 

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