Calculus IV - HW 4 Due 7/20 Section 3.7 1. (Problems 2,3 from B&D) For each of the following, determine !0, R, and δ to write the given expression in the form u = R cos(!0t − δ). p (a) u = − cos(t) + 3 sin(t) (b) u = 4 cos(3t) − 2 sin(3t) Solution: p 2 2 (a) We can immediately see that !0 = 1. We showed in class that R = A + B = q p p p (−1)2 + ( 3)2 = 1 + 3 = 2. Finally, tan(δ) = B = 3 : Thus δ = − π or A −1 p 3 2π δ = 3 . However, we want that 2 cos(δ) = −1 and 2 sin(δ) = 3. Thus we 2π 2π want δ = 3 . Putting this all together, we get u = 2 cos(t − 3 ). p p 2 2 −2 1 (b) Firstly, !0 = 3. Next, R = 4 + 2 = 2 5. Finally tan δ = 4 = − 2 . −1 1 Thus δ = tan (− 2 ) ≈ −:4636 or δ = −:4636 + π. However, we want B = −p2 = sin(δ) < 0 so we choose δ = −:4636. Putting this all together, we R 2 5 p get u = 2 5 cos(3t + 0:4636). 2. (Problems 7 from B&D) A mass weighing 3 lb stretches a spring 3 in. Suppose the mass is pushed upward, contracting the spring a distance of 1 in., and is then set in motion with a downward velocity of 2 ft/s. Assuming there is no damping, find the position u of the mass at any time t. Determine the frequency, period, amplitude, and phase of the motion. 1 ft Solution: To find the mass, we note that w = mg so 3 lb = m · 32 s2 implying 3 lb·s2 m = 32 ft . Since the mass weighing 3 lb stretches the spring 3 in., we have that 1 lb FS = 3 lb = k · 4 ft implying k = 12 ft . Since there is no damping or external force, we have γ = 0 and F (t) = 0. Thus the equation we get is 3 1 u00 + 12u = 0; y(0) = − ; y0(0) = 2 32 12 32 If we multiply through by 3 , we get u00 + 128u = 0 p 2 whichp has characteristic equation r + 128 = 0 whichp has roots rp= ± −128 = ±i · 8 2. Thus the general solution is u(t) = A sin(8 2t) + B cos(8 2t). Plugging in our initial conditions we see y(0) = B = − 1 . Taking the derivative, we get p p p p 12 p y0(t) = A8 2 cos(8 2t) − B8 2 sin(8 2t). In particular y0(0) = A8 2 = 2. Therefore A = p1 . Thus the solution is 4 2 1 p 1 p u(t) = p sin(8 2t) − cos(8 2t): 4 2 12 p 2π π q 1 2 1 2 Here !0 = 8 2 rad=s and T = = p s. As usual, R = p + − = !0 4 2 4 2 12 q q −1=12 1 + 1 = 11 . Finally δ = tan−1 p = π − arctan( p3 ) ≈ 2:0113 since we 32 144 288 1=4 2 2 want sin(δ) > 0 and cos(δ) < 0. 3. (Problems 11 from B&D) A string is stretched 10 cm by a force of 3N. A mass of 2 kg is hung from the spring and is also attached to a viscous damper that exerts a force of 3 N when the velocity of the mass is 5 m/s. If the mass is pulled down 5 cm below its equilibrium position and given an initial downward velocity of 10 cm/s, determine its position u at any time t. Find the quasi frequency µ and the ratio of µ to the natural frequency of the corresponding undamped motion. Solution: We are given m = 2 kg. Furthermore, we know a force of 3N stretches N the spring 10 cm implying FS = 3N = 0:1k cm. Thus k = 30 m . Furthermore, m 3 N·s Fd = 3N = γ · 5 s . Thus γ = 5 m and our equation is 3 2u00 + u0 + 30u = 0; u(0) = :05; u0(0) = 0:1 5 Page 2 Dividing by two we get, 3 u00 + u0 + 15u = 0 10 2 3 which has characteristic equation r + 10 r + 15 = 0 which has roots q 9 3 100 − 4(1)(15) r = − ± = −0:15 ± i3:87008 20 2 Thus the general solution is u = Ae−0:15t sin(3:87008 t) + Be−0:15t cos(3:87008 t): We can plug in our initial conditions and see that u(0) = B = 0:05. Taking the derivative, we get u0 = −0:15Ae−0:15t sin(3:87008 t) + 3:87008Ae−0:15t cos(3:87008 t) −0:15Be−0:15t cos(3:87008 t) − 3:87008Be−0:15t cos(3:87008 t) so u0(0) = 3:87008A − 0:15B = 3:87008A − 0:15(0:05) = 0:1 implying A ≈ 0:02778. Thus the solution is u(t) = 0:02778e−0:15t sin(3:87008 t) + 0:05e−0:15t cos(3:87008 t): Then we see that the quasi frequency is µ = 3:87008 rad=s. Recall that the natural p q k µ 3:87008 frequency is !0 = = 15 rad=s. Thus the ratio is = p ≈ 0:99925. m !0 15 4. (Problems 17 from B&D) A mass weighing 8 lb stretches a spring 1.5 in. The mass is also attached to a damper with coefficient γ. Determine the value of γ for which the system is critically damped; be sure to give the units for γ. Solution: Recall that a system is critically damped when γ2 − 4mk = 0. To 8 lb·s2 convert from weight to mass, we note w = mg so m = 32 ft . For k, we see that the system reaches equilibrium after the spring stretches 1.5 in. Thus Fs +w = 0 so 1:5 1 lb Fs = −w = −8 lb. However, Fs = −kL where here L = 12 = 8 ft. Thus k = 64 ft . Page 3 Thus we have s s p 8 lb · s2 lb lb2 · s2 lb · s γ = 4mk = 4 64 = 64 = 8 : 32 ft ft ft2 ft 5. Optional (Problems 19 from B&D) Assume that the system described by the equation mu00 +γu0 +ku = 0 is either critically damped or overdamped. Show that the mass can pass through the equilibrium position at most once, regardless of the initial conditions. Hint: Determine all possible values of t for which u = 0. Solution: The characteristic equation of mu00 + γu0 + ku = 0 is mr2 + γr + k = 0 with roots γ pγ2 − 4mk r = − ± 2m 2m 2 Since the system is criticallyp damped or overdamped, by definition γ − 4mk ≥ 0. γ γ2−4mk Let α = − 2m and β = 2m . Thus the equation has solutions of the form u(t) = Ae(α+β)t + Be(α−β)t For the mass to pass through the equilibrium point at time t means that u(t) = 0. Thus we set Ae(α+β)t + Be(α−β)t = 0 =) Ae(α+β)t = −Be(α−β)t B =) e(α+β)te−(α−β)t = − A B =) e2βt = − A B B which has exactly one solution if − A > 0 and zero solutions if − A ≤ 0. Thus, the mass can pass through the equilibrium point at most once. Section 3.8 6. (B&D # 5) A mass weighing 4 lb stretches a spring 1.5 in. The mass is displaced 2 in. in the positive direction from its equilibrium position and released with no initial velocity. Assuming that there is no damping and that the mass is acted on by an external force of 2 cos(3t) lb, formulate the initial value problem describing the motion of the mass. Page 4 Solution: We can immediately assume the following. w = 4 lb 1 L = 1:5 in. = 8 ft. s·lb. γ = 0 ft. F (t) = 2 cos(3t) lb 1 u(0) = 2 in. = 6 ft. 0 ft. u (0) = 0 s Some further computation gives w 4 lb 1 lb·s2 m = g ≈ 32 m = 8 m s2 mg 4 lb lb k = L = 1 = 32 ft. 8 ft. It follows that the initial value problem modeling the motion of the mass is 1 1 u00 + 32u = 2 cos(3t); u(0) = ; u0(0) = 0: 8 6 7. (B&D # 6) A mass of 5 kg stretches a spring 10 cm, The mass is acted on by an external force of 10 sin(t=2) N (newtons) and moves in a medium that imparts a viscous force of 2 N when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate the initial value problem describing the motion of the mass. Solution: We can immediately assume the following. m = 5 kg L = 10 cm = :1 m 2 N 2 N kg γ = cm = m = 50 4 s :04 s s F (t) = 10 sin(t=2) N u(0) = 0 m 0 cm m u (0) = 3 s = :03 s Some further computation gives 5 kg·9:8 m mg s2 N k = L ≈ :1 m = 490 m It follows that the initial value problem modeling the motion of the mass is 5u00 + 50u0 + 490u = 10 sin(t=2); u(0) = 0; u0(0) = :03: Page 5 8. (B&D # 9) If an undamped spring-mass system with a mass that weighs 6 lb and a spring constant 1 lb/in is suddenly set in motion at t = 0 by an external force of 4 cos(7t) lb, determine the position of the mass at any time and draw a graph of the displacement vs t.