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gerryliyanaBest ResponseYou've already chosen the best response.5
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 \[x = e^{\frac{ r }{ 2m }} [\frac{ A }{ 2 } e^{i \phi} e^{i \omega t} + \frac{ A }{ 2 } e^{i \phi} e^{i \omega t}]\] \[x = \frac{ A }{ 2 } e^{\frac{ r }{ 2m }t} [e^{i (\omega t + \phi)} + e^{i (\omega t + \phi)}]\]
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.5
@hba @Schrodinger @kryton1212
 one year ago

kryton1212Best ResponseYou've already chosen the best response.0
I am not good at physics.....besides, I am not studying Physics this year... sorry I cannot help you...
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.5
@kryton1212 never mind pretty :D, it's my solution, it's my tutorial about Damped Harmonic Oscillation :D
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.5
@Shadowys Yes, of course, this is a tutorial :D
 one year ago

SchrodingerBest ResponseYou've already chosen the best response.0
^ Proceeds to make things much more awkward than they would ever need to be.
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.5
@Schrodinger show me how you get it done :)
 one year ago

Fools101Best ResponseYou've already chosen the best response.0
That look right to me! *
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.5
thank u @Fools101 :D
 one year ago

Fools101Best ResponseYou've already chosen the best response.0
No problem ^_^ ! so how u been?
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.5
thank u @Gabylovesyou :)
 one year ago

SchrodingerBest ResponseYou've already chosen the best response.0
^ I'll do that as soon as I get through about three more semesters of college, at least, lol.
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.5
ok.., good luck @Schrodinger :)
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
\([C1+C2]=Acos ϕ\) I don't understand why subbing \(ϕ≈\frac{π}{2}\) will result in \([C1+C2]=Acos 0\)
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
Other than that, it seems nice! :)
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.5
nice @Shadowys i mean \[[C_{1}+ C_{2}] = A \cos 90 = 0\] I'm sorry I made a mistake
 one year ago

AperogalicsBest ResponseYou've already chosen the best response.0
@gerryliyana can u tell what partice r u talking about????????
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.5
i'm talking about solving the Damped Harmonic Motion Equation
 one year ago

AperogalicsBest ResponseYou've already chosen the best response.0
So u r writing Transient solution of a general oscillator :) got it.............m i right????????
 one year ago

AperogalicsBest ResponseYou've already chosen the best response.0
@gerryliyana ???????????
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.5
@Aperogalics i'm writing solution of special case at the damped condition of a oscillator
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.5
hei @tyzaQueen hbu?
 one year ago

Hares333Best ResponseYou've already chosen the best response.0
@gerryliyana a tutorial
 one year ago
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