APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS

APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS Second-order linear differential equations have a variety of applications in science and engineering. In this section we explore two of them: the vibration of springs and electric circuits. VIBRATING SPRINGS We consider the motion of an object with mass m at the end of a spring that is either vertical (as in Figure 1) or horizontal on a level surface (as in Figure ). In Section 7.5 we discussed Hooke s Law, which says that if the spring is stretched (or compressed) x units from its natural length, then it exerts a force that is proportional to x: m equilibrium position 0 restoring force kx FIGURE 1 FIGURE x m x equilibrium position m 0 x x where k is a positive constant (called the spring constant). If we ignore any external resisting forces (due to air resistance or friction) then, by Newton s Second Law (force equals mass times acceleration), we have 1 m d x This is a second-order linear differential equation. Its auxiliary equation is mr k 0 with roots r i, where. Thus, the general solution is which can also be written as kx skm or xt c 1 cos t c sin t xt A cost m d x kx 0 where skm (frequency) A sc 1 c (amplitude) cos c 1 A sin c A is the phase angle (See Exercise 17.) This type of motion is called simple harmonic motion. EXAMPLE 1 A spring with a mass of kg has natural length 0.5 m. A force of 5.6 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. SOLUTION From Hooke s Law, the force required to stretch the spring is k0. 5.6 so k 5.60. 18. Using this value of the spring constant k, together with m in Equation 1, we have d x 18x 0 Thomson Brooks-Cole copyright 007 As in the earlier general discussion, the solution of this equation is xt c 1 cos 8t c sin 8t 1

APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS We are given the initial condition that x0 0.. But, from Equation, x0 c 1. Therefore, c 1 0.. Differentiating Equation, we get xt 8c 1 sin 8t 8c cos 8t Since the initial velocity is given as x0 0, we have c 0 and so the solution is xt 1 5 cos 8t DAMPED VIBRATIONS m FIGURE We next consider the motion of a spring that is subject to a frictional force (in the case of the horizontal spring of Figure ) or a damping force (in the case where a vertical spring moves through a fluid as in Figure ). An example is the damping force supplied by a shock absorber in a car or a bicycle. 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.) Thus damping force c dx where c is a positive constant, called the damping constant. Thus, in this case, Newton s Second Law gives or m d x dx restoring force damping force kx c Schwinn Cycling and Fitness m d x dx c kx 0 Equation is a second-order linear differential equation and its auxiliary equation is mr cr k 0. The roots are 4 r 1 c sc 4mk m r c sc 4mk m We need to discuss three cases. x CASE I c 4mk 0 (overdamping) In this case and are distinct real roots and r 1 r x c 1 e r1t c e rt 0 x t Since c, m, and k are all positive, we have sc 4mk c, so the roots r 1 and r given by Equations 4 must both be negative. This shows that x l 0 as t l. Typical graphs of x as a function of t are shown in Figure 4. Notice that oscillations do not occur. (It s possible for the mass to pass through the equilibrium position once, but only once.) This is because c 4mk means that there is a strong damping force (high-viscosity oil or grease) compared with a weak spring or small mass. Thomson Brooks-Cole copyright 007 0 FIGURE 4 Overdamping t CASE II c 4mk 0 (critical damping) This case corresponds to equal roots r 1 r c m

APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS and the solution is given by It is similar to Case I, and typical graphs resemble those in Figure 4 (see Exercise 1), but the damping is just sufficient to suppress vibrations. Any decrease in the viscosity of the fluid leads to the vibrations of the following case. CASE III c 4mk 0 (underdamping) Here the roots are complex: x c 1 c te cmt x r 1 r c m i x=ae (c/m)t where s4mk c m 0 t The solution is given by FIGURE 5 Underdamping x=_ae (c/m)t x e cmt c 1 cos t c sin t We see that there are oscillations that are damped by the factor e cmt. Since c 0 and m 0,wehavecm 0 soe cmt l 0 as t l.this implies that x l 0 as t l ; that is, the motion decays to 0 as time increases. A typical graph is shown in Figure 5. EXAMPLE 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 t if it starts from the equilibrium position and is given a push to start it with an initial velocity of 0.6 ms. SOLUTION From Example 1 the mass is m and the spring constant is k 18, so the differential equation () becomes d x dx 40 18x 0 or d x dx 0 64x 0 The auxiliary equation is r 0r 64 r 4r 16 0 with roots 4 and 16, so the motion is overdamped and the solution is Figure 6 shows the graph of the position function for the overdamped motion in Example. 0.0 xt c 1 e 4t c e 16t We are given that x0 0, so c 1 c 0. Differentiating, we get xt 4c 1 e 4t 16c e 16t so x0 4c 1 16c 0.6 Thomson Brooks-Cole copyright 007 0 1.5 FIGURE 6 Since c c 1, this gives 1c 1 0.6 or c 1 0.05. Therefore x 0.05e 4t e 16t

4 APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS 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 Ft. Then Newton s Second Law gives m d x restoring force damping force external force kx c dx Ft Thus, instead of the homogeneous equation (), the motion of the spring is now governed by the following nonhomogeneous differential equation: 5 m d x dx c kx Ft The motion of the spring can be determined by the methods of Additional Topics: Nonhomogeneous Linear Equations. A commonly occurring type of external force is a periodic force function Ft F 0 cos 0t where 0 skm 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 6 xt c 1 cos t c sin t F 0 m 0 cos 0t If 0, 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). ELECTRIC CIRCUITS switch E R L In Additional Topics: Linear Differential Equations we were able to use first-order linear equations to analyze electric circuits that contain a resistor and inductor. Now that we know how to solve second-order linear equations, we are in a position to analyze the circuit shown in Figure 7. It contains an electromotive force E (supplied by a battery or generator), a resistor R, an inductor L, and a capacitor C, in series. If the charge on the capacitor at time t is Q Qt, then the current is the rate of change of Q with respect to t: I dq. It is known from physics that the voltage drops across the resistor, inductor, and capacitor are FIGURE 7 C RI L di Q C respectively. Kirchhoff s voltage law says that the sum of these voltage drops is equal to the supplied voltage: L di RI Q C Et

APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS 5 Since I dq, this equation becomes 7 L d Q R dq 1 C Q Et which is a second-order linear differential equation with constant coefficients. If the charge and the current are known at time 0, then we have the initial conditions Q 0 I 0 Q0 Q 0 Q0 I0 I 0 and the initial-value problem can be solved by the methods of Additional Topics: Nonhomogeneous Linear Equations. A differential equation for the current can be obtained by differentiating Equation 7 with respect to t and remembering that I dq: L d I di R 1 I Et C EXAMPLE Find the charge and current at time t in the circuit of Figure 7 if R 40, L 1 H, C 16 10 4 F, Et 100 cos 10t, and the initial charge and current are both 0. SOLUTION With the given values of L, R, C, and Et, Equation 7 becomes 8 d Q 40 dq 65Q 100 cos 10t The auxiliary equation is r 40r 65 0 with roots r 40 s900 0 15i so the solution of the complementary equation is Q c t e 0t c 1 cos 15t c sin 15t For the method of undetermined coefficients we try the particular solution Q p t A cos 10t B sin 10t Then Q p t 10A sin 10t 10B cos 10t Q p t 100A cos 10t 100B sin 10t Substituting into Equation 8, we have 100A cos 10t 100B sin 10t 4010A sin 10t 10B cos 10t 65A cos 10t B sin 10t 100 cos 10t or 55A 400B cos 10t 400A 55B sin 10t 100 cos 10t Thomson Brooks-Cole copyright 007 Equating coefficients, we have 55A 400B 100 400A 55B 0 1A 16B 4 or or 16A 1B 0

6 APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS The solution of this system is A 84 and B 64, so a particular solution is and the general solution is Qt Q c t Q p t Imposing the initial condition Q0 0, we get 697 697 Q p t 1 697 84 cos 10t 64 sin 10t e 0t c 1 cos 15t c sin 15t 4 6971 cos 10t 16 sin 10t Q0 c 1 84 697 0 c 1 84 697 To impose the other initial condition we first differentiate to find the current: I dq e 0t 0c 1 15c cos 15t 15c 1 0c sin 15t 40 697 1 sin 10t 16 cos 10t I0 0c 1 15c 640 697 0 c 464 091 Thus, the formula for the charge is Qt 4 e0t 6 cos 15t 116 sin 15t 1 cos 10t 16 sin 10t 697 and the expression for the current is It 1 091 e 0t 190 cos 15t 1,060 sin 15t 101 sin 10t 16 cos 10t 0. Q p Q 0 1. _0. FIGURE 8 NOTE 1 In Example the solution for Qt consists of two parts. Since e 0t l 0 as t l and both cos 15t and sin 15t are bounded functions, So, for large values of t, Q c t 4 091e 0t 6 cos 15t 116 sin 15t l 0 Qt Q p t 4 6971 cos 10t 16 sin 10t as t l and, for this reason, Q p t is called the steady state solution. Figure 8 shows how the graph of the steady state solution compares with the graph of Q in this case. 5 7 m d x L d Q c dx kx Ft R dq 1 Q Et C NOTE Comparing Equations 5 and 7, we see that mathematically they are identical. This suggests the analogies given in the following chart between physical situations that, at first glance, are very different. Spring system Electric circuit x displacement Q charge dx velocity I dq current m mass L inductance c damping constant R resistance k spring constant 1C elastance Ft external force Et electromotive force Thomson Brooks-Cole copyright 007 We can also transfer other ideas from one situation to the other. For instance, the steady state solution discussed in Note 1 makes sense in the spring system. And the phenomenon of resonance in the spring system can be usefully carried over to electric circuits as electrical resonance.

APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS 7 EXERCISES A Click here for answers. 1. A spring with a -kg mass is held stretched 0.6 m beyond its natural length by a force of 0 N. If the spring begins at its equilibrium position but a push gives it an initial velocity of 1. ms, find the position of the mass after t seconds.. A spring with a 4-kg mass has natural length 1 m and is maintained stretched to a length of 1. m by a force of 4. N. If the spring is compressed to a length of 0.8 m and then released with zero velocity, find the position of the mass at any time t.. A spring with a mass of kg has damping constant 14, and a force of 6 N is required to keep the spring stretched 0.5 m beyond its natural length. The spring is stretched 1 m beyond its natural length and then released with zero velocity. Find the position of the mass at any time t. 4. A spring with a mass of kg has damping constant 0 and spring constant 1. (a) Find the position of the mass at time t if it starts at the equilibrium position with a velocity of ms. ; (b) Graph the position function of the mass. 5. For the spring in Exercise, find the mass that would produce critical damping. 6. For the spring in Exercise 4, find the damping constant that would produce critical damping. ; 7. A spring has a mass of 1 kg and its spring constant is k 100. The spring is released at a point 0.1 m above its equilibrium position. Graph the position function for the following values of the damping constant c: 10, 15, 0, 5, 0. What type of damping occurs in each case? ; 8. A spring has a mass of 1 kg and its damping constant is c 10. The spring starts from its equilibrium position with a velocity of 1 ms. Graph the position function for the following values of the spring constant k: 10, 0, 5, 0, 40. What type of damping occurs in each case? 9. Suppose a spring has mass m and spring constant k and let. Suppose that the damping constant is so small that the damping force is negligible. If an external force Ft F 0 cos 0t is applied, where 0, use the method of undetermined coefficients to show that the motion of the mass is described by Equation 6. skm 10. As in Exercise 9, consider a spring with mass m, spring constant k, and damping constant c 0, and let. If an external force Ft F 0 cos t is applied (the applied frequency equals the natural frequency), use the method of undetermined coefficients to show that the motion of the mass is given by xt c 1 cos t c sin t F 0mt sin t. skm 11. Show that if 0, but 0 is a rational number, then the motion described by Equation 6 is periodic. S Click here for solutions. 1. Consider a spring subject to a frictional or damping force. (a) In the critically damped case, the motion is given by x c 1e rt c te rt. Show that the graph of x crosses the t-axis whenever c 1 and c have opposite signs. (b) In the overdamped case, the motion is given by x c 1e r 1t c e r t, where r 1 r. Determine a condition on the relative magnitudes of c 1 and c under which the graph of x crosses the t-axis at a positive value of t. 1. A series circuit consists of a resistor with R 0, an inductor with L 1 H, a capacitor with C 0.00 F, and a 1-V battery. If the initial charge and current are both 0, find the charge and current at time t. 14. A series circuit contains a resistor with R 4, an inductor with L H, a capacitor with C 0.005 F, and a 1-V battery. The initial charge is Q 0.001 C and the initial current is 0. (a) Find the charge and current at time t. ; (b) Graph the charge and current functions. 15. The battery in Exercise 1 is replaced by a generator producing a voltage of Et 1 sin 10t. Find the charge at time t. 16. The battery in Exercise 14 is replaced by a generator producing a voltage of Et 1 sin 10t. (a) Find the charge at time t. ; (b) Graph the charge function. 17. Verify that the solution to Equation 1 can be written in the form xt A cost. 18. The figure shows a pendulum with length L and the angle from the vertical to the pendulum. It can be shown that, as a function of time, satisfies the nonlinear differential equation where t is the acceleration due to gravity. For small values of we can use the linear approximation sin and then the differential equation becomes linear. (a) Find the equation of motion of a pendulum with length 1 m if is initially 0. rad and the initial angular velocity is d 1 rads. (b) What is the maximum angle from the vertical? (c) What is the period of the pendulum (that is, the time to complete one back-and-forth swing)? (d) When will the pendulum first be vertical? (e) What is the angular velocity when the pendulum is vertical? d t L sin 0 L Thomson Brooks-Cole copyright 007

8 APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS ANSWERS S Click here for solutions. 1. x 0.6 sin10t. x 1 5 e 6t 6 5 e t 5. 7. 0.0 c=10 c=15 0 1.4 c=0 c=5 c=0 49 1 kg _0.11 1. Qt e 10t 506 cos 0t sin 0t 15, It 5 e 10t sin 0t 15. Qt e 10t [ 50 cos 0t 500 sin 0t] 50 cos 10t 15 sin 10t Thomson Brooks-Cole copyright 007

APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS 9 SOLUTIONS 1. By Hooke s Law k(0.6) = 0 so k = 100 is the spring constant and the differential equation is x00 + 100 x =0. The general solution is x(t) =c 1 cos 10 t + c sin 10 t.but0=x(0) = c 1 and 1. =x 0 (0) = 10 c,sothe position of the mass after t seconds is x(t) =0.6 sin 10 t.. k(0.5) = 6 or k =1is the spring constant, so the initial-value problem is x 00 +14x 0 +1x =0, x(0) = 1, x 0 (0) = 0. The general solution is x(t) =c 1 e 6t + c e t.but1=x(0) = c 1 + c and 0=x 0 (0) = 6c 1 c. Thus the position is given by x(t) = 1 5 e 6t + 6 5 e t. 5. For critical damping we need c 4mk =0or m = c /(4k) =14 /(4 1) = 49 1 kg. 7. We are given m =1, k = 100, x(0) = 0.1 and x 0 (0) = 0. From (), the differential equation is d x + c dx +100x =0with auxiliary equation r + cr + 100 = 0. Ifc =10,wehavetwocomplexroots r = 5 ± 5 i, so the motion is underdamped and the solution is x = e 5t c 1 cos 5 t + c sin 5 t. Then 0.1 =x(0) = c 1 and 0=x 0 (0) = 5 c 5c 1 c = 1 10,so x = e 5t 0.1cos 5 t 1 10 sin 5 t.ifc =15, we again have underdamping since the auxiliary equation has roots r = 15 ± 5 7 i. The general solution is x = e 15t/ c 1 cos 5 7 t + c sin 0.1 =x (0) = c 1 and 0=x 0 (0) = 5 7 15 c c1 c = 10.Thus 7 x = e 0.1cos 15t/ 5 7 t 10 sin 5 7 t 7. Forc =0, we have equal roots r 1 = r = 10, 5 7 t, so so the oscillation is critically damped and the solution is x =(c 1 + c t)e 10t.Then 0.1 =x(0) = c 1 and 0=x 0 (0) = 10c 1 + c c = 1,sox =( 0.1 t)e 10t.Ifc =5the auxiliary equation has roots r 1 = 5, r = 0, so we have overdamping and the solution is x = c 1 e 5t + c e 0t.Then 0.1 =x(0) = c 1 + c and 0=x 0 (0) = 5c 1 0c c 1 = 15 and c = 1 0, so x = 15 e 5t + 1 0 e 0t.Ifc =0we have roots r = 15 ± 5 5, so the motion is overdamped and the solution is x = c 1 e ( 15 + 5 5 )t + c e ( 15 5 5 )t.then 0.1 =x(0) = c 1 + c and 0=x 0 (0) = 15 + 5 5 c 1 + 15 5 5 c c 1 = 5 5 and c 100 = 5+ 5 x = 5 5 100 100,so e ( 15 + 5 5)t + 5+ 5 100 e ( 15 5 5)t. 9. The differential equation is mx 00 + kx = F 0 cos ω 0t and ω 0 6=ω = k/m. Here the auxiliary equation is mr + k =0with roots ± k/m i = ±ωi so x c (t) =c 1 cos ωt + c sin ωt. Sinceω 0 6=ω,try x p(t) =A cos ω 0t + B sin ω 0t. Thenweneed (m) ω 0 (A cos ω0 t + B sin ω 0 t)+k(a cos ω 0 t + B sin ω 0 t)=f 0 cos ω 0 t or A k mω 0 = F0 and Thomson Brooks-Cole copyright 007 B k mω F 0 F 0 0 =0. Hence B =0and A = = k mω 0 m(ω ω 0 ) since ω = k. Thus the motion of the mass m is given by x(t) =c 1 cos ωt + c sin ωt + F 0 m(ω ω 0 ) cos ω 0t.

10 APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS 11. From Equation 6, x(t) =f(t)+g(t) where f(t) =c 1 cos ωt + c sin ωt and g(t) = cos ω0t. Thenf m(ω ω 0 ) is periodic, with period π ω π,andifω 6=ω0, g is periodic with period ω 0.If ω ω 0 is a rational number, then we can say ω ω 0 = a a = bω b ω 0 where a and b are non-zero integers. Then so x(t) is periodic. x t + a π ω = f t + a π ω = f(t)+g F 0 + g t + a π ω = f(t)+g t + bω ω 0 π ω t + b π ω 0 = f(t)+g(t) =x(t) 1. Here the initial-value problem for the charge is Q 00 +0Q 0 + 500Q =1, Q(0) = Q 0 (0) = 0. Then Q c (t) =e 10t (c 1 cos 0t + c sin 0t) and try Q p (t) =A 500A =1or A = 15. The general solution is Q(t) =e 10t (c 1 cos 0t + c sin 0t)+ 15.But0=Q(0) = c 1 + 15 and Q 0 (t) =I(t) =e 10t [( 10c 1 +0c ) cos 0t +( 10c 0c 1 )sin0t] but 0=Q 0 (0) = 10c 1 +0c.Thus thechargeisq(t) = 1 50 e 10t (6 cos 0t +sin0t)+ 15 and the current is I(t) =e 10t 5 sin 0t. 15. As in Exercise 1, Q c (t) =e 10t (c 1 cos 0t + c sin 0t) but E(t) =1sin10t so try Q p (t) =A cos 10t + B sin 10t. Substituting into the differential equation gives ( 100A +00B + 500A)cos10t +( 100B 00A +500B)sin10t =1sin10t 400A + 00B =0 and 400B 00A =1. Thus A =, B = and the general solution is 50 15 Q(t) =e 10t (c 1 cos 0t + c sin 0t) cos 10t + sin 10t. But0=Q(0) = c 50 15 1 so c 50 1 =. 50 Also Q 0 (t) = 6 sin 10t + cos 10t + 5 5 e 10t [( 10c 1 +0c ) cos 0t +( 10c 0c 1)sin0t] and 0=Q 0 (0) = 6 10c 5 1 +0c so c =. Hence the charge is given by 500 Q(t) =e 10t cos 0t sin 0t cos 10t + sin 10t. 50 500 50 15 c1 17. x(t) =A cos(ωt + δ) x(t) =A[cos ωt cos δ sin ωt sin δ] x(t) =A A cos ωt + c A sin ωt where cos δ = c 1/A and sin δ = c /A x(t) =c 1 cos ωt + c sin ωt. (Notethatcos δ +sin δ =1 c 1 + c = A.) Thomson Brooks-Cole copyright 007