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Simple Harmonic Motion (Total Energy)

  1. Feb 4, 2014 #1
    ImageUploadedByPhysics Forums1391521555.058607.jpg

    I have a question about the derivation that I have attached! I understand that both KE and U are 1/2 kA^2


    So how is it that the two combine is also equal to 1/2kA^2


    Not sure if I'm missing something but I'm a little confused :(
     
  2. jcsd
  3. Feb 4, 2014 #2

    HallsofIvy

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    Part of the problem is that what you "understand" is not true. It is NOT the case that "both KE and U are [itex]1/2 kA^2[/itex]". What is true is that [itex]E= 1/2 kA^2 sin^2(\omega t+ \phi)[/itex] and [itex]U= 1/2 kA^2 cos^2(\omega t+ \phi)[/itex].

    Adding those gives [itex]KE+ U= 1/2 kA^2(sin^2(\theta)+ cos^2(\theta))[/itex]. Now use the fact that [itex]sin^2(\theta)+ cos^2(\theta)= 1[/itex] for all [itex]\theta[/itex]!
     
  4. Feb 4, 2014 #3
    So that leaves

    1/2mv^2 + 1/2kA^2?
     
  5. Feb 4, 2014 #4
    KE = (1/2)mv2 and v= ωAcos(ωt+ø)
    So ,KE = (1/2)mω2A2cos2(ωt+ø)

    But ,k = mω2

    Hence KE = (1/2)kA2cos2(ωt+ø) (1)

    PE = (1/2)kx2 and x= Asin(ωt+ø)
    So ,PE = (1/2)kA2sin2(ωt+ø) (2)

    E=KE+PE

    Using the fact sin2θ+cos2θ =1 ,

    E=(1/2)kA2
     
  6. Feb 4, 2014 #5
    But there are two terms of kA^2?

    I feel like I'm missing something obvious?
     
  7. Feb 4, 2014 #6
    E = (1/2)kA2cos2(ωt+ø) + (1/2)kA2sin2(ωt+ø)

    Take (1/2)kA2 common from both the terms ,

    E = (1/2)kA2[cos2(ωt+ø) + sin2(ωt+ø)]

    Hence E = (1/2)kA2
     
  8. Feb 4, 2014 #7
    Ahh so you take 1/2kA^2 out as a common factor
     
  9. Feb 4, 2014 #8
    Yes...just like if we have 2a+2b ,we can write it as 2(a+b) .
     
  10. Feb 4, 2014 #9
    I knew it would be something trivial! That's a bunch mate!
     
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