Phys105A_W09_HW5sols

Phys105A_W09_HW5sols - Taylor 5.24 As usual when...

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Unformatted text preview: Taylor 5.24 As usual when < we define 1 = 2- 2 and we have a general solution x ( t ) which takes the form of linear combinations of the two independent solutions x 1 ( t ) = e- t cos( 1 t ) (1) x 2 ( t ) = e- t sin( 1 t ) (2) Following Taylors instructions lim x 1 ( t ) lim e- t cos( 1 t ) = lim 1 e- t cos( 1 t ) = e- t Also lim x 2 ( t ) lim e- t sin( 1 t ) = lim 1 e- t sin( 1 t ) = But lim x 2 ( t ) 1 lim e- t sin( 1 t ) 1 = lim 1 e- t sin( 1 t ) 1 = te- t by the well-known trigonometric limit lim x sin( xt ) x = t (3) 1 Taylor 5.27 How many times does x ( t ) = 0? When underdamped, we have the S.H.M solution (which is sinusoidal) multiplied by an exponential decay factor. This crosses x = 0 infinitely many times and so oscil- lator is an appropriate description. (a)When critical damping: x ( t ) = e- t ( C 1 + C 2 t ) . (4) Set this equal to zero and rearrange to find t =- C 1 C 2 (5) This may never be satisfied if t starts from zero and- C 1 C 2 < 0. Oth- erwise it will be satisfied exactly once. (b)When overdamped: We can write the general solution as x ( t ) = e- t ( C 1 e 2- 2 t + C 2 e- 2- 2 t ) = e- t ( C 1 e t + C 2 e- t ) = e- ( - ) t ( C 1 + C 2 e- 2 t ) Set this equal to zero and rearrange to find e- 2 t =- C 1 C 2 (6) The quantity e- 2 t starts at t = 0 as 1, and decreases to 0 when t . At most, there is one time t =- ln(- C 1 /C 2 ) 2 (7) at which x = 0. 2 Taylor 5.36 A = (1000) 2 ((10 ) 2- (2 ) 2 ) 2 +4 ( 2 ) 2 (2 ) 2 ; N [ A ] A = (1000) 2 ((10 ) 2- (2 ) 2 ) 2 +4 ( 2 ) 2 (2 ) 2 ; N [ A ] A = (1000) 2 ((10 ) 2- (2 ) 2 ) 2 +4 ( 2 ) 2 (2 ) 2 ; N [ A ] = ArcTan 2 2 2...
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This note was uploaded on 12/27/2011 for the course PHYS 105a taught by Professor Van,d during the Fall '08 term at UCSB.

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Phys105A_W09_HW5sols - Taylor 5.24 As usual when...

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