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Diff. Eqns. Problem Set 5 Solutions 1. Find the general solution of the nonhomogeneous equation 1 1 y 00 − y 0 + 2 y = −6t2 , t t t > 0. (1) (Hint: One solution of the homogeneous equation is y1 = t.) Since we are given y1 (t), we can use reduction of order to solve this problem. We begin by assuming our solution will be of the form y(t) = v(t)y1 (t) = v(t)t. (2) Plugging this into (1), we get the following equation, and we can solve first for v 0 (t), then integrate to get v(t): 1 1 0 v (t) t + v (t) + 2 v (t) t = −6t2 v 00 (t) t + 2v 0 (t) − t t tv 00 (t) + v 0 (t) = −6t2 d tv 0 (t) = −6t2 dt Z tv 0 (t) = (3) (4) (5) −6t2 dt tv 0 (t) = −2t3 + c1 0 2 (6) (7) −1 v (t) = −2t + c1 t Z v (t) = −2t2 + c1 t−1 dt 2 = − t3 + c1 ln t + c2 . 3 (8) (9) (10) So our solution is y (t) = v (t) y1 (t) 2 3 = − t + c1 ln t + c2 t 3 2 = − t4 + c1 t ln t + c2 t. 3 2. A mass weighing 3 lb contracting the spring a ward velocity of 2 ft/sec of the mass at any time the oscillation. (11) (12) (13) stretches a spring 2 in. If the mass is pushed upward, distance of 2 in, and then set in motion with a downand if there is no damping, then find the position u(t) t. Determine the frequency, period and amplitude of We start with the equation governing this system, mu00 + γu0 + ku = F (t). (14) We can calculate the parameters from the information given. weight 3 lb · s2 3 lb 3 = = = slugs 2 gravity 32 ft 32 32 ft/s damping constant = γ = 0 3 lb lb weight = = 18 spring constant = k = displacement 1/6 ft ft external force = F (t) = 0. mass = m = (15) (16) (17) (18) We are also given initial conditions u(0) = −2 in and u0 (0) = 2 ft/sec, so (14) becomes 3 00 u + 18u = 0, 32 u00 + 192u = 0. u(0) = − 1 ft, 6 u0 (0) = 2 ft/sec (19) (20) √ The characteristic equation is r2 + 192 = 0, with roots r = ±8 3i. Thus our solution is √ √ (21) u (t) = c1 cos 8 3t + c2 sin 8 3t . We use the initial conditions to solve for c1 and c2 . 1 − = u(0) = c1 6 √ 2 = u0 (0) = 8 3c2 (22) 1 ⇒ c2 = √ . 4 3 So our equation for displacement of the spring is √ √ 1 1 u (t) = − cos 8 3t + √ sin 8 3t . 6 4 3 (23) (24) To determine frequency, period and amplitude of the oscillation, we want to reformat our solution from u (t) = A cos (ω0 t) + B sin (ω0 t) (25) u (t) = R cos (ω0 t + δ) , (26) A = R cos δ, B = R sin δ (27) B . A (28) to using the formulas which give us R= p A2 + B 2 , tan δ = √ So in this√case we have A = −1/6, B = 1/(4 3), and the angular frequency of oscillation is ω0 = 8 3 rad/sec. The period can be calculated from the angular frequency as T = 2π π = √ sec. ω0 4 3 (29) Using (27), the amplitude is p R = A2 + B 2 = s 2 √ 1 1 2 7 √ + = − ft. 6 12 4 3 (30) We are not asked to calculate the phase δ. 3. A spring is stretched 5 cm by a force of 2 N. A mass of 2 kg is hung from the spring and is also attached to a viscous damper that exerts a force of 2 N when the velocity of the mass is 4 m/s. If the mass is pulled down 3 cm below its equilibrium position and given an initial upward velocity of 5 cm/s, determine the position u(t) of the mass at any time t. We start with the equation governing this system, mu00 + γu0 + ku = F (t). (31) We can calculate the parameters from the information given. mass = m = 2 kg force 2N 1 N·s 1 kg damping constant = γ = = = = velocity 4 m/s 2 m 2 s weight 2N N kg spring constant = k = = = 40 = 40 2 displacement 0.05 m m s external force = F (t) = 0 N. (32) (33) (34) (35) We can also determine initial conditions from the information. So (31) becomes 1 2u00 + u0 + 40u = 0, 2 u(0) = 0.03 m, u0 (0) = −.05 m/s. (36) The characteristic equation is 2r2 + 12 r + 40 = 0, which has roots r = −1/8 ± The general solution is then ! !! √ √ 1 1279 1279 u (t) = exp − t c1 cos t + c2 sin t . 8 8 8 √ 1279/8i. (37) To find c1 and c2 we use the initial conditions. 0.03 = u(0) = c1 √ −0.05 = u0 (0) = (38) 1279 1 c2 − c1 8 8 ⇒ c2 = − 37 √ ≈ −0.010346. 100 1279 (39) Thus our solution is 1 u (t) = exp − t 8 √ 0.03 cos ! 1279 37 √ t − sin 8 100 1279 √ 1279 t 8 !! (40) 4. The displacement u(t) of a mass-spring-damper system is governed by the equation mu00 + γu0 + ku = F0 cos ωt (41) where m is the mass, γ is the damping constant, k is the spring constant, F0 is the amplitude of the applied force, and ω is the frequency of the periodic forcing. (a) Consider first the undamped case with γ = 0. Find the value of ω for which resonance occurs for the case m = 2, k = 10 and F0 = 3. Resonance occurs when the natural frequency ωp 0 equals ω, √ the frequency of the applied force.√ The natural frequency is given by ω0 = k/m = 5. So resonance occurs when ω = 5. (b) Now consider a damped system with γ = 1. Determine the amplitude of the forced oscillation for the case m = 3, k = 2, F0 = 1 and ω = 1. (Hint: See page 208.) Page 208 gives us the following formula for the amplitude of the forced oscillation: F0 R= q , 2 m2 ω02 − ω 2 + γ 2 ω 2 ω02 = k/m. (42) First calculate ω02 = k/m = 2/3. Then 1 1 R= q =√ . 2 32 (2/3 − 12 )2 + 12 · 12 (43)