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gerryliyana
 3 years ago
Problem Solving  Damped Harmonic Motion
gerryliyana
 3 years ago
Problem Solving  Damped Harmonic Motion

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gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.6Damped 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 \[x = e^{\frac{ r }{ 2m }} [\frac{ A }{ 2 } e^{i \phi} e^{i \omega t} + \frac{ A }{ 2 } e^{i \phi} e^{i \omega t}]\] \[x = \frac{ A }{ 2 } e^{\frac{ r }{ 2m }t} [e^{i (\omega t + \phi)} + e^{i (\omega t + \phi)}]\]

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.6@hba @Schrodinger @kryton1212

kryton1212
 3 years ago
Best ResponseYou've already chosen the best response.0I am not good at physics.....besides, I am not studying Physics this year... sorry I cannot help you...

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.6@kryton1212 never mind pretty :D, it's my solution, it's my tutorial about Damped Harmonic Oscillation :D

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.6@Shadowys Yes, of course, this is a tutorial :D

Schrodinger
 3 years ago
Best ResponseYou've already chosen the best response.0^ Proceeds to make things much more awkward than they would ever need to be.

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.6@Schrodinger show me how you get it done :)

Fools101
 3 years ago
Best ResponseYou've already chosen the best response.0That look right to me! *

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.6thank u @Fools101 :D

Fools101
 3 years ago
Best ResponseYou've already chosen the best response.0No problem ^_^ ! so how u been?

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.6thank u @Gabylovesyou :)

Schrodinger
 3 years ago
Best 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.

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.6ok.., good luck @Schrodinger :)

Shadowys
 3 years ago
Best 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 \([C1+C2]=Acos 0\)

Shadowys
 3 years ago
Best ResponseYou've already chosen the best response.1Other than that, it seems nice! :)

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.6nice @Shadowys i mean \[[C_{1}+ C_{2}] = A \cos 90 = 0\] I'm sorry I made a mistake

Aperogalics
 3 years ago
Best ResponseYou've already chosen the best response.0@gerryliyana can u tell what partice r u talking about????????

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.6i'm talking about solving the Damped Harmonic Motion Equation

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

Aperogalics
 3 years ago
Best ResponseYou've already chosen the best response.0@gerryliyana ???????????

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.6@Aperogalics i'm writing solution of special case at the damped condition of a oscillator

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.6hei @tyzaQueen hbu?

Hares333
 3 years ago
Best ResponseYou've already chosen the best response.0@gerryliyana a tutorial

gerryliyana
 10 months ago
Best ResponseYou've already chosen the best response.6thank you @perl :)
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