1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

SHM Equation problem

  1. Jun 7, 2005 #1

    DDS

    User Avatar

    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
     
  2. jcsd
  3. Jun 7, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You need to find where the rate of change of PE is maximum, not where the PE is maximum.
     
  4. Jun 7, 2005 #3

    DDS

    User Avatar

    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??
     
  5. Jun 7, 2005 #4

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  6. Jun 7, 2005 #5

    Doc Al

    User Avatar

    Staff: Mentor

    You were on the right track. If you write the potential energy ([itex]1/2 k x^2[/itex]) 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 in to 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.)
     
  7. Jun 7, 2005 #6

    DDS

    User Avatar

    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?
     
  8. Jun 7, 2005 #7

    DDS

    User Avatar

    Is there someone who can help me from here?
     
  9. Jun 7, 2005 #8

    Doc Al

    User Avatar

    Staff: Mentor

    Show how you obtained this result.
     
  10. Jun 7, 2005 #9

    DDS

    User Avatar

    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?
     
  11. Jun 7, 2005 #10

    Doc Al

    User Avatar

    Staff: Mentor

    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.
     
  12. Jun 7, 2005 #11

    DDS

    User Avatar

    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>?
     
  13. Jun 7, 2005 #12

    Doc Al

    User Avatar

    Staff: Mentor

    Not for nothing. It is a necessary step, you just stopped too soon.

    Hint: KE = Total Energy - PE

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

    Can you write the PE as a function of t?
     
  14. Jun 7, 2005 #13

    DDS

    User Avatar

    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 dont know how to manipoulate it to get to where i have to go
     
  15. Jun 7, 2005 #14

    Doc Al

    User Avatar

    Staff: Mentor

    Well... what possessed you to square the function given and take its derivative?

    Hint: Spring potential energy is given by [itex]1/2 k x^2[/itex], where x is the displacement from equilibrium. You are given x(t). Use it.
     
  16. Jun 7, 2005 #15

    DDS

    User Avatar

    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 doesnt make sense to me seeing as how we are looking for time when the change is the greatest
     
  17. Jun 7, 2005 #16

    Doc Al

    User Avatar

    Staff: Mentor

    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.)
     
  18. Jun 7, 2005 #17

    DDS

    User Avatar

    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
     
  19. Jun 8, 2005 #18

    Doc Al

    User Avatar

    Staff: Mentor

    You want an expression for spring PE:
    [itex]\mbox{PE} = 1/2 k x^2 = 1/2 k A^2 \cos^2 \omega t[/itex]
    Realize that the derivative of [itex]\cos^2 \omega t[/itex] is [itex]-2 \omega \cos \omega t \sin \omega t[/itex], not [itex]2 \cos \omega t - \sin \omega t[/itex]. (I assume you are just writing it wrongly?)

    See my correction above.

    Once you get the correct expression for d(PE)/dt, then yes, -d(PE)/dt will equal d(KE)/dt.

    You find the maxima of that function just like any other.
     
    Last edited: Jun 8, 2005
  20. Jun 8, 2005 #19

    DDS

    User Avatar

    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 im a littel consfused as to how i would find the maxima of such a function
     
  21. Jun 8, 2005 #20

    Doc Al

    User Avatar

    Staff: Mentor

    As I stated in the last post, the formula for PE is: [itex]\mbox{PE} = 1/2 k x^2 = 1/2 k A^2 \cos^2 \omega t[/itex]

    Thus:
    [itex]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[/itex]
    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.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: SHM Equation problem
  1. SHM problem (Replies: 2)

  2. SHM Problem (Replies: 8)

  3. A SHM problem! (Replies: 7)

  4. A SHM problem (Replies: 4)

  5. SHM Problem (Replies: 1)

Loading...