1 Difference Equations to Differential Equations Section 8.5 Applications: Pendulums Mass-Spring Systems In this section we will investigate two applications of our work in Section 8.4. First, we will consider the motion of a pendulum, a prolem originally mentioned in Section. in connection with the trigonometric functions. Second, we will discuss the motion of an oject virating at the end of a spring. The motion of a pendulum Consider a pendulum consisting of a o of mass m at the end of a rigid rod of length. We will assume that the mass of the rod is negligile in comparison with the mass of the o. Let x(t e the angle etween the rod the vertical at time t, with x(t > 0 for angles measured in the counterclockwise direction x(t < 0 for angles measured in the clockwise direction. See Figure Suppose the o is pulled through an angle α then released. That is, suppose our initial conditions are x(0 = α ẋ(0 = 0. If we view the motion of the pendulum in the complex plane, with the real axis vertical, positive direction downward, the imaginary axis horizontal, positive direction to the right, then the position of the o at time t is given y z(t = e ix(t. (8.5.1 x(t Figure A pendulum Then we have ż = iẋe ix (8.5. z = ẋ e ix + iẍe ix = ẋ (cos(x + i sin(x + iẍ(cos(x + i sin(x (8.5.3 = ( ẋ cos(x ẍ sin(x + i( ẋ sin(x + ẍ cos(x. 1 Copyright c y Dan Sloughter 000
2 Applications: Pendulums Mass-Spring Systems Section 8.5 Now z is the acceleration of the pendulum, so m z must e equal to the force of gravity acting on the o, namely, a force of magnitude mg acting in the downward direction, the direction of the positive real axis. Hence we must have g = z, that is, g = ( ẋ cos(x ẍ sin(x + i( ẋ sin(x + ẍ cos(x. (8.5.4 Equating the real imaginary parts of the two sides of (8.5.4 gives us g = ẋ cos(x ẍ sin(x ( = ẋ sin(x + ẍ cos(x. (8.5.6 Multiplying (8.5.5 y sin(x (8.5.6 y cos(x gives us g sin(x = ẋ cos(x sin(x + ẍ sin (x ( = ẋ sin(x cos(x + ẍ cos (x. (8.5.8 Adding (8.5.7 (8.5.8 together yields g sin(x = ẍ(sin (x + cos (x = ẍ. (8.5.9 Thus ẍ = g sin(x. ( So we have reduced the prolem of descriing the motion of the pendulum to the prolem of solving the second order differential equation ( suject to the initial conditions x(0 = α ẋ(0 = 0. Unfortunately, this equation is not linear. In fact, it is not possile to find a closed form solution for this equation. In Section 8.6 we will discuss how to study this equation using numerical approximations, ut for now we will take a different approach to finding an approximate solution. Since we know sin(x = x + o(x ( from our work on est affine approximations in Chapter, it is reasonale to replace sin(x y x for small values of x. Hence, if we restrict to the case where α is small, we may replace ( y the linear equation ẍ = g x. (8.5.1 Since this equation is homogeneous with constant coefficients, we may solve it using the techniques of Section 8.4. Specifically, the characteristic equation for this equation is k + g = 0, (8.5.13
3 Section 8.5 Applications: Pendulums Mass-Spring Systems Figure 8.5. Motion of a pendulum which has roots Hence the general solution is Then g k 1 = i g k = i ( ( ( g ( x = c 1 cos t g + c sin t. ( ẋ = c 1 g sin ( g t + c g cos ( g t, ( so x(0 = c 1 ẋ(0 = c g. Hence the initial conditions x(0 = α ẋ(0 = 0 imply c 1 = α c = 0. Thus ( g x = α cos t ( The graph of x for the case = 1 meter α = 0.1 radians, in which case we use g = 9.8 meters per second per second, is shown in Figure One consequence of ( is that the period of the motion, that is, the time it takes the o to make one complete oscillation, is π g = π g, ( independent of the value of α. Of course, we are working under the approximation sin(x x, so ( is actually only an approximation of the period. Nevertheless,
4 4 Applications: Pendulums Mass-Spring Systems Section 8.5 the approximation is very good for small oscillations is the reason pendulums were used to measure time in early clocks. Virations in mechanical systems: mass-spring systems In this example we consider the motion of an oject of mass m suspended on a spring, as shown in Figure We will measure the position of the oject along a vertical axis, with the equilirium position at 0 the positive direction downward. Let x(t denote the position of the oject at time t suppose the oject is released from rest at position x 0. That is, we suppose that x(0 = x 0 ẋ(0 = 0. If we ignore any damping forces, such as resistance to the motion due to the surrounding medium, such as air or oil, then the only forces acting on the oject are the force of gravity, contriuting a term of mg, the restorative force of the spring, given, according to Hooke s law, y kl for some constant k > 0, where l is the amount the spring is stretched or compressed from its natural length. If we let l e the amount the spring is stretched when the oject is at the equilirium position, that is, when x = 0, then at any time the spring is stretched or compressed y x + l. Thus at any time t the force acting on the oject is F = mg k(x + l. (8.5.0 x = 0 Figure Mass on a spring at equilirium In particular, if the oject is at rest at its equilirium position, then oth x = 0 F = 0. Hence 0 = mg k l, (8.5.1 so mg = k l. (8.5. Thus (8.5.0 simplifies to F = kx. Applying Newton s second law of motion, we have mẍ = kx, (8.5.3 from which we otain ẍ = k x. (8.5.4 m
5 Section 8.5 Applications: Pendulums Mass-Spring Systems Figure Motion of a mass-spring system without damping This equation is of the same form as the equation derived aove for approximating the motion of a pendulum. Hence, using the same reasoning, the solution is ( k x = x 0 cos m t. (8.5.5 The graph of x for k = 10, m = 5, x 0 = is shown in Figure Notice that the period of the motion is T = π k m m = π k. (8.5.6 The frequency of the motion, that is, the numer of complete oscillations in one unit of time, is f = 1 T = 1 k π m. (8.5.7 Hence for a fixed mass, increasing the spring constant, that is, increasing the stiffness of the spring, decreases the period increases the frequency; for a fixed spring constant, increasing the mass increases the period decreases the frequency. Now suppose there is a damping force, a force resisting the motion of the oject, which is proportional to the velocity. This adds an additional term of cẋ, where c is a positive constant, to the force acting on the oject, giving us F = kx cẋ. Thus mẍ = kx cẋ, (8.5.8 so ẍ + c mẋ + k m x = 0 (8.5.9
6 6 Applications: Pendulums Mass-Spring Systems Section 8.5 replaces (8.5.4 as the equation descriing the motion of the oject. notation, we will let = c m Then our differential equation ecomes a = k m. To simplify the ẍ + ẋ + a x = 0, ( with characteristic equation (using s for the variale s + s + a = 0. ( Hence the roots of the characteristic equation are s 1 = 4 4a = a (8.5.3 s = + 4 4a = + a. ( Thus the ehavior of the system depends on whether a > 0, a = 0, or a < 0. Equivalently, since a = c 4m k m, the ehavior of the system depends on whether c > 4mk, c = 4mk, or c < 4mk. In the first case the system is said to e overdamped, in the second it is critically damped, in the third it is underdamped. First consider the overdamped case a > 0. In this case the characteristic equation has distinct real roots, so the general solution is x = c 1 e s 1t + c e s t. ( Now ẋ = c 1 s 1 e s 1t + c s e s t, ( so x(0 = c 1 + c ẋ(0 = c 1 s 1 + c s. Hence the initial conditions, x(0 = x 0 ẋ(0 = 0, give us x 0 = c 1 + c 0 = c 1 s 1 + c s.
7 Section 8.5 Applications: Pendulums Mass-Spring Systems Figure Motion of an overdamped mass-spring system Multiplying the first equation y s 1 sutracting from the second gives us Hence Thus Now > 0 > a, so Hence x 0 s 1 = c (s s 1. c = x 0s 1 s s 1 c 1 = x 0 c = x 0(s s 1 s s 1 + x 0s 1 s s 1 = x 0s s s 1. It follows that e s t > e s 1t, s s 1 > 0, x = x 0 s s 1 (s e s 1t s 1 e s t. ( s = + a < 0. s 1 < s < 0. ( s e s 1t s 1 e s t > s e s t s 1 e s t = e s t (s s 1 > 0 for all t 0. Hence if x 0 < 0, then x(t < 0 for all t 0, if x 0 > 0, then x(t > 0 for all t > 0. Comining this with lim x(t = 0, ( t we see that in this case the system does not oscillate at all. After release, the oject simply returns to the equilirium position. Figure shows this ehavior for k = 10, m = 5, c = 0, x 0 =.
8 8 Applications: Pendulums Mass-Spring Systems Section Figure Motion of a critically damped mass-spring system Next consider the case when a = 0. In this case the characteristic equation has only one real root, s 1 = s =, so the general solution is x = c 1 e t + c te t. ( Then ẋ = c 1 e t c te t + c e t, ( so x(0 = c 1 ẋ(0 = c 1 + c. Hence the initial conditions, x(0 = x 0 ẋ(0 = 0, give us c 1 = x 0 c = x 0. Thus Equivalently, since = Now for any t 0, x = x 0 e t + x 0 te t = x 0 e t (1 + t. ( c m, x = x 0 e c m t( c m t > 0. c m t. (8.5.4 Hence, as in the overdamped case, the system does not oscillate. Once released, the oject moves ack to the equilirium position without ever crossing it. Figure shows this ehavior for k = 10, m = 5, c = 10, x 0 =. This motion is said to e critically damped ecause any increase in c results in overdamped motion, while any decrease in c results in underdamped motion, which we consider next. Finally, consider the case when a < 0. The roots of the characteristic equation are now s 1 = a = i a ( s = + a = + i a (8.5.44
9 Section 8.5 Applications: Pendulums Mass-Spring Systems 9 If we let α = a, then the general solution is Then x = e t (c 1 cos(αt + c sin(αt. ( ẋ = e t ( αc 1 sin(αt + αc cos(αt e t (c 1 cos(αt + c sin(αt, ( so x(0 = c 1 ẋ(0 = αc c 1. Hence the initial conditions, x(0 = x 0 ẋ(0 = 0, imply that c 1 = x 0 Thus c = x 0 α. x = e t (x 0 cos(αt + x 0 α sin(αt = x 0 α e t (α cos(αt + sin(αt. ( This expression simplifies somewhat if we introduce the angle Then ( θ = tan 1. ( α cos(θ = α α + sin(θ = α +. Moreover, since α = a, Hence x = x 0 α + α α + = (a + = a = k m. ( e t α α + cos(αt + = x 0 k α m e t (cos(θ cos(αt + sin(θ sin(αt. α + sin(αt Using the angle sutraction formula for cosine, this ecomes x = x 0 k α m e t cos(αt θ. ( The presence of the cosine factor in this expression shows us that, even though we still have lim x(t = 0, t
10 10 Applications: Pendulums Mass-Spring Systems Section Figure Motion of an underdamped mass-spring system the underdamped mass-spring system will oscillate aout the equilirium position with a decreasing amplitude of x 0 k α m e t. ( Figure shows this ehavior for k = 10, m = 5, c = 5, x 0 =. Prolems 1. In an experiment to determine g, a pendulum of length 50 centimeters is oserved to have a period of oscillation of 1.4 seconds. Approximate g ased on this oservation.. The period of oscillation of a pendulum of length given in ( is, as mentioned, only an approximation of the true period. It can e shown that the true period of a pendulum released from an angle α is given y T = 4 g π 0 1 dφ, 1 k sin (φ where 0 < α < π k = sin ( α. (a Find the period of oscillation for a pendulum of length 50 centimeters for α = π 4, α = π 6, α = π 50, α = π 100. Compare these results with the approximation given in ( ( Graph T as a function of α for π 4 α π 4. For comparison, also plot the horizontal line T = π g. 3. Consider a mass-spring system with x 0 = 10, ẋ(0 = 0, k = 10, m = 10. Plot x(t for c = 0, c = 5, c = 10, c = 0, c = 5, c = 30. Identify each motion as overdamped, critically damped, underdamped, or undamped.
11 Section 8.5 Applications: Pendulums Mass-Spring Systems Consider a mass-spring system with x 0 = 10, ẋ(0 = 0, m = 10, c = 0. Plot x(t for k =, k = 5, k = 10, k = 15. Identify each motion as overdamped, critically damped, underdamped, or undamped. 5. Consider the underdamped motion of a mass-spring system expressed in ( (a Show that the maximum values of x(t occur at t = 0, T, T,..., where T = π. k m c 4m Note that when c = 0, T reduces to the period of the motion for the mass-spring system without damping. ( Show that if x 1 x are two successive maximum values of x(t, then x 1 x = e ct m. 6. Inside the earth, the force of gravity acting on an oject is proportional to the distance etween the oject the center of the earth. (a Suppose a hole is drilled through the earth from pole to pole a rock is dropped into the hole. If x(t is the distance from the oject to the center of the earth at time t, show that, ignoring any resistive forces, where R is the radius of the earth. ( g x = R cos R t, ( How long, in minutes, does it take for the rock to make one complete trip from pole to pole ack? Use R = 3950 miles. (c What is the velocity of the rock, in miles per hour, when it reaches the center of the earth?