P2214HW2 - PHYS2214, Spring 2009 Solutions to Assignment 2...

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PHYS2214, Spring 2009 Solutions to Assignment 2 February 11, 2009 1 Free oscillations a) Recall that the damped oscillator considered in lecture satisﬁes, m d 2 x dt 2 + b dx dt + kx = 0 (1) If we assume a solution of the form z ( t ) = Ae αt (2) where α is in general complex, and plug into eqn.[1], we get 2 + + = 0 (3) the solutions of which are α ± = - b ± b 2 - 4 mk 2 m (4) = - 1 τ A ± s 1 τ 2 A - ω 2 0 (5) where τ A = 2 m b is the amplitude decay time and ω 0 = q k m is the natural frequency in the absence of damping. We evaluate the solution in the underdamped, overdamped and critically damped case with initial conditions x ( t = 0) = A and v ( t = 0) = 0. 1) Underdamped case: ω 0 τ A > 1 1 τ 2 A - ω 2 0 < 0 α ± = - 1 τ A ± i q ω 2 0 - 1 τ 2 A ≈ - 1 τ A ± 0 z ( t ) = Be α + t + Ce α - t (6) x ( t ) = Re [ z ( t )] = Be - t τ A cos( ω 0 t + φ ) (7) Imposing x ( t = 0) = A and v ( t = 0) = 0 gives, Ae - t τ A cos ω 0 t (8) 2) Critically damped case: ω 0 τ A = 1 α = - 1 τ A x ( t ) = Re [ z ( t )] = Ae - t τ A (9) 1

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satisﬁes the initial conditions. Notice that in the critically damped case, there is only 1 solution to eqn.[3]. If you are wondering why there is only one constant of integration, check that te αt is also a solution and hence the most general solution is x ( t ) = ( B + Ct ) e αt . The iniitial conditions are satisﬁed for B = A and C = 0. 3) Overdamped case: ω 0 τ A < 1 1 τ 2 A - ω 2 0 > 0 α ± = - 1 τ A ± q 1 τ 2 A - ω 2 0 z ( t ) = ˜ Be α + t + ˜ Ce α - t (10) x ( t ) = Be ( - 1 τ A + q 1 τ 2 A - ω 2 0 ) t + Ce ( - 1 τ A - q 1 τ 2 A - ω 2 0 ) t (11) Notice that both α + and α - are real and negative and that B and C are the real parts of ˜ B and ˜ C . x ( t = 0) = A (12) B + C = A (13) v ( t = 0) = 0 (14) + + - = 0 (15) which on solving gives B = A 1 τ A + q 1 τ 2 A - ω 2 0 2 q 1 τ A 2 - ω 2 0 (16) C = A - 1 τ A + q 1 τ 2 A - ω 2 0 2 q 1 τ A 2 - ω 2 0 (17) Fig.[1] shows a plot of x ( t ) vs t for the 3 cases above. We take A = 1 and ω = 1 and τ = 100, 1 and 0 . 01 respectively. (b) In the underdamped case the frequency of oscillation, ω = ω 0 s 1 - 1 ( ω 0 τ A ) 2 (18) The values of ω 0 τ A required to produce a particular shift in the natural frequency is given by, ω 0 + δω = ω 0 s 1 - 1 ( ω 0 τ A ) 2 (19) ± 1 + δω ω 0 ² 2 = ± 1 - 1 ( ω 0 τ A ) 2 ² (20) 1 ( ω 0 τ A ) 2 = - 2 δω ω 0 + ± δω 0 ω 0 ² 2 (21) ω 0 τ A = 1 q 2 | δω | ω 0 + ( δω ω 0 ) 2 (22) We now calculate the values of ω 0 τ required to produce various percentage changes in natural frequency. (1) 0.1 percent : ω 0 τ = 22 . 36 2
(2) 1 percent : ω 0 τ = 7 . 05 (3) 10 percent : ω 0 τ = 2 . 18 (c) (1) Auto suspension system : Critically damped: towrds underdamped side (2) Door bell : Underdamped (3) Speaker Cone : Underdamped (4) Pogo stick : Underdamped (5) Drum head : Critical/Under (6) Cymbal : Underdamped (7) Bungee cord+person : Underdamped 2 Driven oscillations a)The complex amplitude of the responce of a driven oscillator described by

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This note was uploaded on 04/01/2009 for the course PHYS 2214 taught by Professor Giambattista,a during the Spring '07 term at Cornell University (Engineering School).

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P2214HW2 - PHYS2214, Spring 2009 Solutions to Assignment 2...

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