Phys105A_W09_HW5sols

# Phys105A_W09_HW5sols - Taylor 5.24 As usual when β< ω...

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Unformatted text preview: Taylor 5.24 As usual when β < ω we define ω 1 = ω 2- β 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- βt cos( ω 1 t ) (1) x 2 ( t ) = e- βt sin( ω 1 t ) (2) Following Taylor’s instructions lim β → ω x 1 ( t ) lim β → ω e- βt cos( ω 1 t ) = lim ω 1 → e- βt cos( ω 1 t ) = e- βt Also lim β → ω x 2 ( t ) lim β → ω e- βt sin( ω 1 t ) = lim ω 1 → e- βt sin( ω 1 t ) = But lim β → ω x 2 ( t ) ω 1 lim β → ω e- βt sin( ω 1 t ) ω 1 = lim ω 1 → e- βt sin( ω 1 t ) ω 1 = te- βt 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- βt ( 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- βt ( C 1 e √ β 2- ω 2 t + C 2 e- √ β 2- ω 2 t ) = e- βt ( C 1 e λt + C 2 e- λt ) = e- ( β- λ ) t ( C 1 + C 2 e- 2 λt ) Set this equal to zero and rearrange to find e- 2 λt =- C 1 C 2 (6) The quantity e- 2 λt 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 λ (7) at which x = 0. 2 Taylor 5.36 A = (1000) 2 ((10 π ) 2- (2 π ) 2 ) 2 +4 ( π 2 ) 2 (2 π ) 2 ; N [ A ] A = (1000) 2 ((10 π ) 2- (2 π ) 2 ) 2 +4 ( π 2 ) 2 (2 π ) 2 ; N [ A ] A = (1000) 2 ((10 π ) 2- (2 π ) 2 ) 2 +4 ( π 2 ) 2 (2 π ) 2 ; N [ A ] δ = ArcTan 2 π 2 2...
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Phys105A_W09_HW5sols - Taylor 5.24 As usual when β< ω...

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