{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Phys 2214 hw1 Solutions

# Phys 2214 hw1 Solutions - Phys 2214 Homework#1 Solutions...

This preview shows pages 1–3. Sign up to view the full content.

Phys 2214 Homework #1 Solutions September 1, 2011 1. z ( t ) = re i ( ωt + φ ) (a) z ( t ) = re i ( ωt + φ ) = r [cos( ωt + φ ) + i sin( ωt + φ )]. Then x ( t ) = Re[ z ( t )] = r cos( ωt + φ ) and y ( t ) = Im[ z ( t )] = r sin( ωt + φ ) . (b) Suppose ω = ω 0 + 1 . Then z ( t ) = re i ( ωt + φ ) = re i ( ω 0 + 1 t + φ ) = re - ω 1 t e i ( ω 0 t + φ ) and Re[ z ( t )] = re - ω 1 t cos( ω 0 t + φ ). (c) i. dz dt = iωre i ( ωt + φ ) ii. d dt { Re[ z ( t )] } = d dt [ r cos( ωt + φ )] = - ωr sin( ωt + φ ) iii. Re " dz dt # = Re[ iωre i ( ωt + φ ) ] = Re { iωr [cos( ωt + φ ) + i sin( ωt + φ )] } = Re[ iωr cos( ωt + φ ) - ωr sin( ωt + φ )] = - ωr sin( ωt + φ ) By comparing (ii) with (iii), we see that d dt [ ] and Re[ ] com- mute. (d) Given the equation a d 2 x dt 2 + b dx dt + cx = 0 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
If x ( t ) = Re[ z ( t )], then if we substitute this into the equation above and use the result from part (b) we get: Re { a d 2 z dt 2 + b dz dt + cz } = 0 . In general we can’t conclude that if Re[ z ( t )] = 0 then z ( t ) = 0, since it isn’t necessary that Im[ z ( t )] = 0. However, in this case, substitute z ( t ) = re i ( ωt + φ ) into the previous equation to get Re[ z ( t )( - 2 + ibω + c )] = 0 . Since z ( t ) contains the only time dependence, the only way for this equation to hold is if - 2 + ibω + c = 0. But if we substitute z ( t ) = re i ( ωt + φ )
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}