{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solution_1_08_

# Solution_1_08_ - 1 Physics 214 Spring 2008 Solutions to...

This preview shows pages 1–2. Sign up to view the full content.

1 Physics 214 Spring 2008 Solutions to Assignment ] 1 Problem A.1. ( a ). Given z = re i ( ωt + φ ) with real r , ω , and φ , we have Re [ z ] = r cos( ωt + φ ) and Im [ z ] = r sin( ωt + φ ). ( b ). Now z = re i (( ω 0 + 1 ) t + φ ) = re - ω 1 t e i ( ω 0 t + φ ) . Therefore, Re [ z ] = re - ω 1 t cos( ωt + φ ) and Im [ z ] = re - ω 1 t sin( ωt + φ ). ( c ). (i). dz/dt = iωz , (ii). dx/dt = d/dtRe [ z ] = - ωr sin( ωt + φ ), (ii). Re [ dz/dt ] = - ωr sin( ωt + φ ). ( d ). With x = Re [ z ] and a ¨ z + b ˙ z + cz = ( - 2 + ibω + c ) z = 0, we have ( - 2 + ibω + c ) = 0 which gives ω = i b 2 a ± b 2 a p - 1 + 4 ac/b 2 . Assuming a , b and c are real, we have Re [ ω ] = b 2 a p - 1 + 4 ac/b 2 and Im [ ω ] = b 2 a for 4 ac/b 2 > 1. For 4 ac/b 2 < 1, we have Re [ ω ] = 0 and Im [ ω ] = b 2 a ± b 2 a p 1 - 4 ac/b 2 . For 4 ac/b 2 = 1, we have Re [ ω ] = 0 and Im [ ω ] = b 2 a . Problem A.2. Solving e = cos φ + i sin φ and e - = cos φ - i sin φ for cos φ and sin φ gives cos φ = ( e + e - ) / 2 and sin φ = ( e - e - ) / 2 i . Problem A.3. (a). 1 200 1 100 3 200 1 50 t LParen1 s RParen1 1 Minus 1 Minus 2 Minus 3 Minus 4 Minus 5 Minus 6 Minus 7 x LParen1 cm RParen1 Figure 1: Plot of x ( t ) = x eq + A cos[ ωt + φ 0 ] for x eq = - 3 cm , A = 4

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

Solution_1_08_ - 1 Physics 214 Spring 2008 Solutions to...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online