Solving for Oscillation with Given Parameters

  • Thread starter Thread starter skrat
  • Start date Start date
  • Tags Tags
    Oscillation
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 1K views
skrat
Messages
740
Reaction score
8

Homework Statement


Solve: ##\ddot{x}+\Omega^{2} x=D+\frac{C}{2}+Ecos\omega t+\frac{C}{2}cos2\omega t##


Homework Equations





The Attempt at a Solution


I got a hint to use ##x=\alpha sin\omega t+\beta cos\omega t## so ##\ddot{x}=-\alpha ^{2}\omega ^{2}sin\omega t-\beta ^{2}\omega ^{2}cos\omega t## in the equation above than:
##(-\alpha ^{2}\omega ^{2}sin\omega t-\beta ^{2}\omega ^{2}cos\omega t)+\Omega^{2}x=\alpha sin\omega t+\beta cos\omega t=D+\frac{C}{2}+Ecos\omega t+\frac{C}{2}cos2\omega t##
Which gives me 4 separate equations depending on ##sin\omega t##, ##cos\omega t##, ##cos2\omega t## and constant:

first: ##-\alpha ^{2}\omega ^{2}+\Omega ^{2}\alpha=0##
second: ##-\beta ^{2}\omega ^{2}+\Omega ^{2}\beta =E##
third: ##\frac{C}{2}=0##
fourth: ##D+\frac{C}{2}=0##

Forth and third together say that ##D=0## and ##C=0##
First says that:
##\alpha ^{2}\omega ^{2}=\Omega ^{2}\alpha##
##\alpha =(\frac{\Omega }{\omega })^{2}##
But for second I am not sure, whether I can divide it with ##\beta## (probably not since it could be equal to 0) or how do I solve it?
PLEASE HELP

BTW, if everything is completely wrong and this is not how usually this kind of equations are solved, please let me know.
 
Physics news on Phys.org
C and D are given, so if you deduce values for them you have gone wrong. Seems to me you need more flexibility in your proposed solution, probably terms involving sin and cos of 2ωt.