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SECOND-ORDER DIFFERENTIAL EQUATIONS

14. SECOND-ORDER DIFFERENTIAL EQUATIONS. SECOND-ORDER DIFFERENTIAL EQUATIONS. Second-order linear differential equations have a variety of applications in science and engineering. SECOND-ORDER DIFFERENTIAL EQUATIONS. 14.3 Applications of Second-Order Differential Equations.

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SECOND-ORDER DIFFERENTIAL EQUATIONS

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  1. 14 SECOND-ORDER DIFFERENTIAL EQUATIONS

  2. SECOND-ORDER DIFFERENTIAL EQUATIONS • Second-order linear differential equations have a variety of applications in science and engineering.

  3. SECOND-ORDER DIFFERENTIAL EQUATIONS 14.3 Applications of Second-Order Differential Equations In this section, we will learn how: Second-order differential equations are applied to the vibration of springs and electric circuits.

  4. VIBRATING SPRINGS • We consider the motion of an object with mass m at the end of a spring that is either vertical or horizontal on a level surface.

  5. SPRING CONSTANT • In Section 6.4, we discussed Hooke’s Law: • If the spring is stretched (or compressed) x units from its natural length, it exerts a force that is proportional to x: restoring force = –kxwhere k is a positive constant, called the spring constant.

  6. SPRING CONSTANT Equation 1 • If we ignore any external resisting forces (due to air resistance or friction) then, by Newton’s Second Law, we have: • This is a second-order linear differential equation.

  7. SPRING CONSTANT • Its auxiliary equation is mr2 + k = 0 with roots r = ±ωi, where • Thus, the general solution is: x(t) = c1 cos ωt + c2 sin ωt

  8. SIMPLE HARMONIC MOTION • This can also be written as x(t) = A cos(ωt + δ) • where: • This is called simple harmonic motion.

  9. VIBRATING SPRINGS Example 1 • A spring with a mass of 2 kg has natural length 0.5 m. • A force of 25.6 N is required to maintain it stretched to a length of 0.7 m. • If the spring is stretched to a length of 0.7 m and then released with initial velocity 0, find the position of the mass at any time t.

  10. VIBRATING SPRINGS Example 1 • From Hooke’s Law, the force required to stretch the spring is: k(0.25) = 25.6 • Hence, k = 25.6/0.2 = 128

  11. VIBRATING SPRINGS Example 1 • Using that value of the spring constant k, together with m = 2 in Equation 1, we have:

  12. VIBRATING SPRINGS E. g. 1—Equation 2 • As in the earlier general discussion, the solution of the equation is: x(t) = c1 cos 8t + c2 sin 8t

  13. VIBRATING SPRINGS Example 1 • We are given the initial condition: x(0) = 0.2 • However, from Equation 2, x(0) = c1 • Therefore, c1 = 0.2

  14. VIBRATING SPRINGS Example 1 • Differentiating Equation 2, we get: x’(t) = –8c1 sin 8t + 8c2 cos 8t • Since the initial velocity is given as x’(0) = 0, we have c2 = 0. • So, the solution is: x(t) = (1/5) cos 8t

  15. DAMPED VIBRATIONS • Now, we consider the motion of a spring that is subject to either: • A frictional force (the horizontal spring here) • A damping force (where a vertical spring moves through a fluid, as here)

  16. DAMPING FORCE • An example is the damping force supplied by a shock absorber in a car or a bicycle.

  17. DAMPING FORCE • We assume that the damping force is proportional to the velocity of the mass and acts in the direction opposite to the motion. • This has been confirmed, at least approximately, by some physical experiments.

  18. DAMPING CONSTANT • Thus, • where c is a positive constant, called the damping constant.

  19. DAMPED VIBRATIONS Equation 3 • Thus, in this case, Newton’s Second Law gives: • or

  20. DAMPED VIBRATIONS Equation 4 • Equation 3 is a second-order linear differential equation. • Its auxiliary equation is: mr2 + cr + k = 0

  21. DAMPED VIBRATIONS Equation 4 • The roots are: • According to Section 13.1, we need to discuss three cases.

  22. CASE I—OVERDAMPING • c2 – 4mk > 0 • r1 and r2 are distinct real roots. • x = c1er1t + c2er2t

  23. CASE I—OVERDAMPING • Since c, m, and k are all positive, we have: • So, the roots r1 and r2 given by Equations 4 must both be negative. • This shows that x→ 0 as t→ ∞.

  24. CASE I—OVERDAMPING • Typical graphs of x as a function of fare shown. • Notice that oscillations do not occur. • It’s possible for the mass to pass through the equilibrium position once, but only once.

  25. CASE I—OVERDAMPING • This is because c2 > 4mk means that there is a strong damping force (high-viscosity oil or grease) compared with a weak spring or small mass.

  26. CASE II—CRITICAL DAMPING • c2 – 4mk = 0 • This case corresponds to equal roots • The solution is given by: x = (c1 + c2t)e–(c/2m)t

  27. CASE II—CRITICAL DAMPING • It is similar to Case I, and typical graphs resemble those in the previous figure. • Still, the damping is just sufficient to suppress vibrations. • Any decrease in the viscosity of the fluid leads to the vibrations of the following case.

  28. CASE III—UNDERDAMPING • c2 – 4mk < 0 • Here, the roots are complex: where • The solution is given by: x =e–(c/2m)t(c1 cos ωt +c2 sin ωt)

  29. CASE III—UNDERDAMPING • We see that there are oscillations that are damped by the factor e–(c/2m)t. • Since c > 0 and m > 0, we have –(c/2m) < 0. So, e–(c/2m)t→ 0 as t→ ∞. • This implies that x→ 0 as t→∞. That is, the motion decays to 0 as time increases.

  30. CASE III—UNDERDAMPING • A typical graph is shown.

  31. DAMPED VIBRATIONS Example 2 • Suppose that the spring of Example 1 is immersed in a fluid with damping constant c = 40. • Find the position of the mass at any time tif it starts from the equilibrium position and is given a push to start it with an initial velocity of 0.6 m/s.

  32. DAMPED VIBRATIONS Example 2 • From Example 1, the mass is m = 2 and the spring constant is k = 128. • So, the differential equation 3 becomes:or

  33. DAMPED VIBRATIONS Example 2 • The auxiliary equation is: • r2 + 20r + 64 = (r + 4)(r + 16) = 0 • with roots –4 and –12. • So, the motion is overdamped, and the solution is: x(t) = c1e–4t + c2e–16t

  34. DAMPED VIBRATIONS Example 2 • We are given that x(0) = 0. • So, c1 + c2 = 0. • Differentiating, we get:x’(t) = –4c1e–4t – 16c2e–16t • Thus, x’(0) = –4c1 – 16c2 = 0.6

  35. DAMPED VIBRATIONS Example 2 • Since c2 = –c1, this gives: • 12c1 = 0.6 or c1 = 0.05 • Therefore, x = 0.05(e–4t – e–16t)

  36. FORCED VIBRATIONS • Suppose that, in addition to the restoring force and the damping force, the motion of the spring is affected by an external force F(t).

  37. FORCED VIBRATIONS • Then, Newton’s Second Law gives:

  38. FORCED VIBRATIONS Equation 5 • So, instead of the homogeneous equation 3, the motion of the spring is now governed by the following non-homogeneous differential equation: • The motion of the spring can be determined by the methods of Section 13.2

  39. PERIOD FORCE FUNCTION • A commonly occurring type of external force is a periodic force function F(t) = F0 cos ω0t • where ω0 ≠ ω =

  40. PERIOD FORCE FUNCTION Equation 6 • In this case, and in the absence of a damping force (c = 0), you are asked in Exercise 9 to use the method of undetermined coefficients to show that:

  41. RESONANCE • If ω0 = ω, then the applied frequency reinforces the natural frequency and the result is vibrations of large amplitude. • This is the phenomenon of resonance. • See Exercise 10.

  42. ELECTRIC CIRCUITS • In Sections 9.3 and 9.5, we were able to use first-order separable and linear equations to analyze electric circuits that contain a resistor and inductor or a resistor and capacitor.

  43. ELECTRIC CIRCUITS • Now that we know how to solve second-order linear equations, we are in a position to analyze this circuit.

  44. ELECTRIC CIRCUITS • It contains in series: • An electromotive force E(supplied by a battery or generator) • A resistor R • An inductor L • A capacitor C

  45. ELECTRIC CIRCUITS • If the charge on the capacitor at time tis Q = Q(t), then the current is the rate of change of Q with respect to t: I = dQ/dt

  46. ELECTRIC CIRCUITS • As in Section 9.5, it is known from physics that the voltage drops across the resistor, inductor, and capacitor, respectively, are:

  47. ELECTRIC CIRCUITS • Kirchhoff’s voltage law says that the sum of these voltage drops is equal to the supplied voltage:

  48. ELECTRIC CIRCUITS Equation 7 • Since I = dQ/dt, the equation becomes: • This is a second-order linear differential equation with constant coefficients.

  49. ELECTRIC CIRCUITS • If the charge Q0 and the current I0 are known at time 0, then we have the initial conditions: • Q(0) = Q0Q’(0) = I(0) = I0 • Then, the initial-value problem can be solved by the methods of Section 13.2

  50. ELECTRIC CIRCUITS • A differential equation for the current can be obtained by differentiating Equation 7 with respect to t and remembering that I = dQ/dt:

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