Assignment_2solns - Solutions to Assignment 2 Physics 214...

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Solutions to Assignment 2 Physics 214, Spring 2008 February 8, 2008 Problem 1: Damped Oscillators In the previous assignment we studied the damped oscillator in the under- damped regime. In this problem we look at the general case. The nature of the solution depends on the dimensionless parameter ω 0 τ . (a) Free Oscillations The equation of motion is m d 2 x dt 2 + b dx dt + kx = 0 (1) As usual we obtain the solution of eqn.[1] by finding the real part of a complex function z ( t ) that satisfies it. This is done so by substituting z ( t ) = e αt and solving for α . 2 + + 2 0 = 0 α ± = - b ± b 2 - 4 km 2 m = - 1 τ ± r 1 τ 2 - ω 2 0 (2) as τ = 2 m b . 1) Underdamped case; ω 0 τ > 1 1 τ 2 - ω 2 0 < 0 α ± = - 1 τ ± i q ω 2 0 - 1 τ 2 z ( t ) = Be α + t + Ce α - t x ( t ) = Re [ z ( t )] = Ae - t τ cos( r ω 2 0 - 1 τ 2 t ) (3) 2) Critically damped case; ω 0 τ = 1 α = - 1 τ x ( t ) = Re [ z ( t )] = Ae - t τ (4) 1
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3) Overdamped case; ω 0 τ < 1 1 τ 2 - ω 2 0 > 0 α ± = - 1 τ ± q 1 τ 2 - ω 2 0 z ( t ) = Ce α + t + De α - t x ( t ) = Ae ( - 1 τ + p 1 τ 2 - ω 2 0 ) + Be ( - 1 τ - p 1 τ 2 - ω 2 0 ) (5) Notice that both α + and α - are real and negative and that A and B are the real parts of C and D . (b) Plots of x(t) for the 3 damping regimes: (See end of document) (c) The frequency of the oscillations in the underdamped case; ω = ω 0 q ω 2 0 - 1 ( ω 0 τ ) 2 We now calculate the values of ω 0 τ required to produce various percentage changes in natural frequency. ω 0 + δω = ω 0 s 1 - 1 ( ω 0 ) τ ) 2 (1 + δω ω 0 ) 2 = (1 - 1 ( ω 0 τ ) 2 ) 1 ( ω 0 τ ) 2 = - 2 δω ω 0 + ( δω 0 ω 0 ) 2 ω 0 τ = 1 q 2 | δω | ω 0 + ( δω ω 0 ) 2 (6) We note above that δω is negative. We find using Eqn. [6] that the required ω 0 τ
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