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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 wellknown 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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 Fall '08
 Van,D
 Sin, Cos

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