anonymous
  • anonymous
Problem Solving - Damped Harmonic Motion
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Is this a tutorial?
anonymous
  • anonymous
Damped SHM \[x = e ^{\frac{-r }{ 2m }t} (C _{1} e ^{i \omega t} + C_{2} e^{- \omega t}) \] a) when t = 0, then: \[x = A \cos \phi \] where: \[\frac{ dx }{ dt } = - \omega A \sin \phi \] approachable \[\frac{ r }{ m } \approx 0\] \[\phi = \frac{ \pi }{ 2 }\] calculate the value of C1 and C2! MY SOLUTION when t = 0 \[x(t) = e^{-\frac{ r }{ 2m }(t)} (C_{1} e^{i \omega (t)} + C_{2} e^{- i \omega (t)})\] \[x (0) = e ^{-\frac{ r }{ 2m } (0)} (C_{1} e^{i \omega (0)} + C_{2} e^{-i \omega (0)})\] \[x(0) = e^{0} (C_{1} e^{0} + C_{2}e^{0})\] \[x (0) = (C_{1}+ C_{2})\] because of when t=0, x = a cos phi \[x(0) = [C_{1}+ C{2}] = A \cos \phi\] \[[C_{1}+C_{2}] = A \cos \phi ........ (1)\] the next section \[\frac{ dx }{ dt } = \frac{ d }{ dt }(e^{-\frac{ r }{ 2m }t} [C_{1}e^{i \omega t} +C_{2}e^{-i \omega t}])\] \[\frac{ d }{ dt }(e^{i \omega t - \frac{ r }{ 2m }t} C_{1}) + \frac{ d }{ dt }(e^{-i \omega t - \frac{ r }{ 2m}t} C_{2})\] \[e^{i \omega t - \frac{ r }{ 2m }t} (i \omega - \frac{ r }{ 2m }) C_{1} + e^{-i \omega t - \frac{ r }{ 2m}t}(-i \omega -\frac{ r }{ 2m })\] when t = 0 \[\frac{ dx }{ dt } = e^{0} (i \omega - \frac{ r }{ 2m })C_{1} + e^{-0} (-i \omega - \frac{ r }{ 2m })\] because of : \[\frac{ dx }{ dt } = - \omega A \sin \phi \] Then \[e^{i \omega (0) - \frac{ r }{ 2m }(0)} (i \omega - \frac{ r }{ 2m }) C_{1} + e^{-i \omega (0) - \frac{ r }{ 2m}(0)}(-i \omega -\frac{ r }{ 2m })= - \omega A \sin \phi\] \[i \omega (C_{1} - C_{2}) - \frac{ r }{ 2m }(C_{1}+C{2}) = -\omega A \sin \phi\] when approachable \[\frac{ r }{ m } \approx 0\] then \[i \omega (C_{1} - C_{2}) -(0)(C_{1}+C{2})) = -\omega A \sin \phi\] \[i \omega (C_{1} -C_{2}) = - \omega A \sin \phi ......(2)\] when approachable \[\phi \approx \frac{ \pi }{ 2 } \] then based on a second (2) equation \[[C_{1}+C_{2}] = A \cos \phi\] \[[C_{1}+C_{2}] = A \cos 0\] \[[C_{1}+ C_{2}] \approx 0\] then \[C_{1}- C_{2} = -\frac{ A \sin \phi }{ i } = i A \sin \phi\] \[C_{1}+C_{2} = A \cos \phi ..... (3) \] \[C_{1}-C_{2} = i A \sin \phi ..... (4)\] for C1, add (3) and (4) \[(C_{1}+C_{2}) +(C_{1}-C_{2}) = A \cos \phi + i A \sin \phi \] \[(C_{1}+C_{1}) + (C_{2}-C_{2}) = A \cos \phi + i A \sin \phi\] \[2C_{1} = A \cos \phi + i A \sin \phi\] \[C_{1}= \frac{ A }{ 2 } (\cos \phi + i \sin \phi)\] \[C_{1} = \frac{ A }{ 2 } e ^{i \phi}\] for C2, subtract (3) and (4) \[(C_{1}+C_{2}) - (C_{1}- C_{2}) = A \cos \phi - i A \sin \phi\] \[C_{1}+C_{2} -C_{1} + C_{2} = A \cos \phi - i A \sin \phi\] \[2 C_{2} = A \cos \phi - i A \sin \phi\] \[C_{2} = \frac{ A }{ 2 } e^{-i \phi} \] Finally obtained the value of C1 and C2 \[C_{1} = \frac{ A }{ 2 } e ^{i \phi}\] \[C_{2} = \frac{ A }{ 2 } e^{-i \phi} \] The new equation of Damped Harmonic Oscillation \[x = e^{-\frac{ r }{ 2m }} [\frac{ A }{ 2 } e^{i \phi} e^{i \omega t} + \frac{ A }{ 2 } e^{-i \phi} e^{-i \omega t}]\] \[x = \frac{ A }{ 2 } e^{-\frac{ r }{ 2m }t} [e^{i (\omega t + \phi)} + e^{-i (\omega t + \phi)}]\]
anonymous
  • anonymous
@hba @Schrodinger @kryton1212

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More answers

anonymous
  • anonymous
I am not good at physics.....besides, I am not studying Physics this year... sorry I cannot help you...
anonymous
  • anonymous
@kryton1212 never mind pretty :D, it's my solution, it's my tutorial about Damped Harmonic Oscillation :D
anonymous
  • anonymous
@Shadowys Yes, of course, this is a tutorial :D
anonymous
  • anonymous
errrr........
anonymous
  • anonymous
@UnkleRhaukus
Schrodinger
  • Schrodinger
^ Proceeds to make things much more awkward than they would ever need to be.
anonymous
  • anonymous
@Schrodinger show me how you get it done :)
Fools101
  • Fools101
That look right to me! *
anonymous
  • anonymous
thank u @Fools101 :D
Fools101
  • Fools101
No problem ^_^ ! so how u been?
Gabylovesyou
  • Gabylovesyou
correct :)
anonymous
  • anonymous
thank u @Gabylovesyou :)
Schrodinger
  • Schrodinger
^ I'll do that as soon as I get through about three more semesters of college, at least, lol.
anonymous
  • anonymous
ok.., good luck @Schrodinger :)
hba
  • hba
good going.
anonymous
  • anonymous
\([C1+C2]=Acos ϕ\) I don't understand why subbing \(ϕ≈\frac{π}{2}\) will result in \([C1+C2]=Acos 0\)
anonymous
  • anonymous
Other than that, it seems nice! :)
anonymous
  • anonymous
nice @Shadowys i mean \[[C_{1}+ C_{2}] = A \cos 90 = 0\] I'm sorry I made a mistake
anonymous
  • anonymous
@gerryliyana can u tell what partice r u talking about????????
anonymous
  • anonymous
i'm talking about solving the Damped Harmonic Motion Equation
anonymous
  • anonymous
So u r writing Transient solution of a general oscillator :) got it.............m i right????????
anonymous
  • anonymous
@gerryliyana ???????????
anonymous
  • anonymous
@Aperogalics i'm writing solution of special case at the damped condition of a oscillator
anonymous
  • anonymous
hei @tyzaQueen hbu?
anonymous
  • anonymous
@gerryliyana a tutorial
perl
  • perl
nice work
anonymous
  • anonymous
thank you @perl :)

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