anonymous 3 years ago Problem Solving - Damped Harmonic Motion

1. anonymous

Is this a tutorial?

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

3. anonymous

@hba @Schrodinger @kryton1212

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

errrr........

8. anonymous

@UnkleRhaukus

9. anonymous

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

10. anonymous

@Schrodinger show me how you get it done :)

11. Fools101

That look right to me! *

12. anonymous

thank u @Fools101 :D

13. Fools101

No problem ^_^ ! so how u been?

14. Gabylovesyou

correct :)

15. anonymous

thank u @Gabylovesyou :)

16. anonymous

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

17. anonymous

ok.., good luck @Schrodinger :)

18. hba

good going.

19. anonymous

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

20. anonymous

Other than that, it seems nice! :)

21. anonymous

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

22. anonymous

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

23. anonymous

i'm talking about solving the Damped Harmonic Motion Equation

24. anonymous

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

25. anonymous

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

26. anonymous

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

27. anonymous

hei @tyzaQueen hbu?

28. anonymous

@gerryliyana a tutorial

29. perl

nice work

30. anonymous

thank you @perl :)