Phys 2214 hw2 Solutions

# Phys 2214 hw2 Solutions - Phys 2214 Homework#2 Solutions...

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Unformatted text preview: Phys 2214 Homework #2 Solutions September 9, 2011 1. Free oscillations (a) For the initial conditions x (0) = A and v (0) = 0, we have the following results: i. Underdamped: x ( t ) = Ae- t/τ A cos ( ω t ) ii. Overdamped: x ( t ) = Ae- t/τ A iii. Critically damped: x ( t ) = A (1 + t/τ A ) e- t/τ A These are plotted in the figure. Critically Damped Underdamped Overdamped 2 4 6 8 10- 1.0- 0.5 0.0 0.5 1.0 (b) For the underdamped case, since b 2 4 mk , α =- b ± √ b 2- 4 mk 2 m 1 becomes α =- b ± i √ 4 mk- b 2 2 m . The imaginary part of α is Im( α ) = √ 4 mk- b 2 2 m = q ω 2- 1 /τ 2 A . Since ω decreases when ω τ A 6 = 0, write ω-| Δ ω | = ω q 1- 1 / ( ω τ A ) 2 . Dividing by ω , squaring both sides, and solving for ω τ A gives ω τ A = 1 q 2 | Δ ω | /ω- ( | Δ ω | /ω ) 2 . i. For | Δ ω | /ω = 0 . 001, ω τ A = 22 . 4. ii. For | Δ ω | /ω = 0 . 01, ω τ A = 7 . 09. iii. | Δ ω | /ω = 0 . 1, ω τ A = 2 . 29. (c) For the underdamped case, the system undergoes many oscilla- tions during the time it takes to relax to its equilibrium position. For the critically damped and the overdamped cases, the system doesn’t undergo any oscillations. i. Auto suspension: critically damped; slightly underdamped. ii. Door bell: underdamped. iii. Speaker cone: underdamped. iv. Pogo stick: underdamped. v. Drum head: critical or underdamped. vi. Cymbal: underdamped. vii. Bungee cord + person: underdamped. 2. Driven oscillations A ( ω D ) = F /m ω 2- ω 2 D + i 2 ω D /τ A (a) To find the magnitude, use | A ( ω D ) | = q A * ( ω D ) A ( ω D ) = F m 1 q ( ω 2- ω 2 D ) 2 + 4 ω 2 D /τ 2 A 2 To find the phase, rewrite A ( ω D ) as A ( ω D ) = F m 1 ω 2- ω 2 D + i 2 ω D /τ A × ω 2- ω 2 D- i 2 ω D /τ A ω 2- ω 2 D- i 2 ω D /τ A = F m ω 2- ω 2 D- i 2...
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Phys 2214 hw2 Solutions - Phys 2214 Homework#2 Solutions...

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