Question

Problem Solving - Damped Harmonic Motion

Answer 1

Is this a tutorial?

Answer 2

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
\[x = e^{-\frac{ r }{ 2m }} [\frac{ A }{ 2 } e^{i \phi} e^{i \omega t} + \frac{ A }{ 2 } e^{-i \phi} e^{-i \omega t}]\]
\[x = \frac{ A }{ 2 } e^{-\frac{ r }{ 2m }t} [e^{i (\omega t + \phi)} + e^{-i (\omega t + \phi)}]\]

Answer 3

@hba @Schrodinger @kryton1212

Answer 4

I am not good at physics.....besides, I am not studying Physics this year...
sorry I cannot help you...

Answer 5

@kryton1212 never mind pretty :D, it's my solution, it's my tutorial about Damped Harmonic Oscillation :D

Answer 6

@Shadowys Yes, of course, this is a tutorial :D

Answer 7

errrr........

Answer 8

@UnkleRhaukus

Answer 9

^ Proceeds to make things much more awkward than they would ever need to be.

Answer 10

@Schrodinger show me how you get it done :)

Answer 11

That look right to me! *

Answer 12

thank u @Fools101 :D

Answer 13

No problem ^_^ ! so how u been?

Answer 14

correct :)

Answer 15

thank u @Gabylovesyou :)

Answer 16

^ I'll do that as soon as I get through about three more semesters of college, at least, lol.

Answer 17

ok.., good luck @Schrodinger :)

Answer 18

good going.

Answer 19

\([C1+C2]=Acos ϕ\)
I don't understand why subbing \(ϕ≈\frac{π}{2}\) will result in \([C1+C2]=Acos 0\)

Answer 20

Other than that, it seems nice! :)

Answer 21

nice @Shadowys
i mean
\[[C_{1}+ C_{2}] = A \cos 90 = 0\]
I'm sorry I made a mistake

Answer 22

@gerryliyana can u tell what partice r u talking about????????

Answer 23

i'm talking about solving the Damped Harmonic Motion Equation

Answer 24

So u r writing Transient solution of a general oscillator :)
got it.............m i right????????

Answer 25

@gerryliyana ???????????

Answer 26

@Aperogalics i'm writing solution of special case at the damped condition of a oscillator

Answer 27

hei @tyzaQueen hbu?

Answer 28

@gerryliyana a tutorial

Answer 29

nice work

Answer 30

thank you @perl :)