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Unformatted text preview: Sample Problems for Midterm #2  Solutions 1) Given: m = 10 kg, k = 250 N/m, m ¨ x + c ˙ x + kx = F ( t ) a) ω = q k m = s 250 N/m 10 kg = 5 rad/s b) β = c 2 m = 3 rad/s β 2 ω 2 = (3 rad/s) 2 (5 rad/s) 2 = (4 rad/s) 2 < ⇒ under damped Q = ω r 2 β = 1 2 r ω β 2 2 = 1 2 q 25 9 2 = √ 7 6 ≈ . 441 critically damped if c = 2 mβ = c crit = 2 mω = 2 · 10 kg · 5 rad/s = 100 kg/s c) Given: F ( t ) = F cos( ω f t ) with F = 48 N and driving frequency ω f . ’steady state’: = particular solution x p ( t ) = D ( ω f ) cos( ω f t δ ( ω f )) where D ( ω f ) = F /m √ ( ω 2 ω 2 f ) 2 +(2 βω f ) 2 and δ ( ω f ) = tan 1 2 βω f ω 2 ω 2 f . ’maximum amplitude’: = resonance: ω r = √ ω 2 2 β 2 = √ 7 rad/s. resonance amplitude D r = D ( ω r ) = 48 N / 10 kg q ( (5) 2 ( √ 7) 2 ) 2 + ( 2 · 3 · √ 7 ) 2 rad/s 2 = 0 . 2m resonance phase difference δ r = δ ( ω r ) = tan 1 2 · 3 · √ 7 5 2 √ 7 2 ( rad/s ) 2 ( rad/s ) 2 = tan 1 ( √ 7 / 3) ≈ 41 . 4 ◦ 2) a) m ¨ x = F tot = F damp + F harm + F driv = c ˙ x mω 2 x + θ ( t ) F ( t ) Thus ¨ x + 2 β ˙ x + ω 2 x = θ ( t ) f ( t ), with β = c/ 2 m = ω/ 10, Heaviside function θ ( t ) and f ( t ) = 1 m ( A sin( ωt ) + B sin(3 ωt )), is the differential equation we would like to solve, subject to the initial conditions that the mass is initially at rest: x (∞ ) = ˙ x (∞ ) = 0. Since this harmonic oscillator is under damped: β 2 ω 2 = . 99 ω 2 < we have x h ( t ) = C 1 e ωt/ 10 cos( √ . 99 ωt ) + C 2 e ωt/ 10 sin( √ . 99 ωt ) as the general homogeneous solution. The driving force starts up at t = 0. We are therefore effectively having a homogeneous differential equation for t < 0 with solution of form x < ( t ) = x h ( t ) and an inhomogeneous differential equation for t > 0 with solution of type...
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This note was uploaded on 02/23/2009 for the course PHYS 325 taught by Professor Lamb during the Fall '09 term at University of Illinois at Urbana–Champaign.
 Fall '09
 Lamb
 mechanics

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