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PHYS310_homework_04_solutions

# PHYS310_homework_04_solutions - Homework set 4 due...

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Unformatted text preview: Homework set 4, due Wednesday, September 21, 11:30 am Following problems from Analytical Mechanics by Fowles & Cassiday 1. 3.2 (20 points) m & x = 0.1ωo cos ωot x = 0.1sin ω ot [m] s m & x = 0.5 = 0.1ω o When t = 0, x = 0 and s 2π T= = 1.26 s ωo = 5 s −1 ωo 2. 3.5 (20 points) 1 2121 212 & & mx1 + kx1 = mx2 + kx2 2 2 2 2 2 2 &2 & k ( x1 − x2 ) = m ( x2 − x12 ) 1 &2 & k x2 − x12 2 ωo = = 2 2 m x1 − x2 121 212 & kA = mx1 + kx1 2 2 2 & & m x2 x2 − x2 x2 & A2 = x12 + x12 = 1 12 2 1 + x12 2 & & k x2 − x1 1 2 x12 x2 − x2 x12 2 &2 & A= 2 2 & & x2 − x1 3. 3.10 (20 points) c k = 3 s −1 ωo2 = = 25 s −2 (a) γ = 2m m 2 2 2 −2 ω d = ω o − γ = 16 s ω r 2 = ω d 2 − γ 2 = 7 s −2 (b) (c) ∴ω r = 7 s −1 F 48 Amax = o = m = 0.2 m Cωd 60.4 tan φ = 2γω r 2γω r ω r 7 = = = 2 2 2 ( ωo − ω r ) 2γ γ 3 4. 3.15 (20 points) Amaxγ A( ω ) ≈ 1 ( ω o − ω ) 2 + γ 2 2 ∴φ ≈ 41.4o 1 for A ( ω ) = Amax , 2 γ 1 = 2 1 ( ω o − ω ) 2 + γ 2 2 2 2 2 ( ωo − ω ) + γ = 4γ ωo − ω = ±γ 3 ω = ωo ± γ 3 5. 3.18 (20 points) βt Using the hint, Fext = Re ( Foe ) , where β = −α + iω , and x(t) is the real part of the solution to: && & mx + cx + kx = Foe β t . Assuming a solution of the form: x = Ae β t −iφ ( mβ 2 + cβ + k ) x = Fo xeiφ A F mα 2 − 2imαω − mω 2 − cα + icω + k = o ( cos φ + i sin φ ) A F m ( α 2 − ω 2 ) − cα + k = o cos φ A Fo ω ( −2mα + c ) = sin φ A ω ( c − 2mα ) φ = tan −1 m ( α 2 − ω 2 ) − cα + k Using sin 2 φ + cos 2 φ = 1 , 2 Fo2 2 = m ( α 2 − ω 2 ) − cα + k + ω 2 ( c − 2mα ) 2 A Fo A= { m ( α 2 − ω 2 ) − cα + k + ω 2 ( c − 2mα ) 2 2 −α t and x ( t ) = Ae cos ( ω t − φ ) + the transient term. } 1 2 ...
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PHYS310_homework_04_solutions - Homework set 4 due...

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