Statement The front shock absorber of a typical mountain bike (see Figure 1) may be mod- eled as a spring-dashpot system. A typical value for the spring constant might be k = 15000 newtons per meter with a damping constant c = 1700 newtons per meter per second. The mass in this system consists of the rider's mass and the bike's mass, less the mass of the wheels since they are not suspended by the shock absorber. Suppose the rider has a mass of 80 kg, the bike (less wheels) has a mass of 12 kg, and that half of this total mass is supported by the front shock absorber, so the effective supported mass is m = (80 + 12)/2 = 46 kg. Assume that the only other force acting on the rider is gravity. If the front wheel is in contact with the ground, it may be considered fixed or immovable, and if u(f) denotes the vertical displacement of the front shock from equilibrium then ma"(t) + ex'(t) + ku(t) = f(t) becomes 46u" (1) + 1700u' (t) + 15000u(t) = -450.8. (1) where f(f) = -my = 450.8 with g =9.8, half the weight of the bike and rider. Modeling Exercise 1 Suppose the rider has been pedaling on level ground for some distance, so it's reasonable to assume that w(f) has settled to a constant or equilibrium value, u(t) = Me, for some constant weg. To find me, substitute u(t) = u into the ODE (1) and use the fact that u'(t) = u"(f) =0 here. How far does the shock absorber compress under the rider's weight? A typical bike front shock has a range of motion of about 140 mm before bottoming out, and it is recommended that the rider's weight alone should compress the shock 20 to 30 percent of the shock's range of motion. Is this recommendation satisfied here?Modeling Exercise 2 Find the general solution to (1). Let's use this result to do some practical analysis. Suppose the rider of this mountain bike rides off a ledge or jump that's 1.5 meters in height. In the air there is no force on the shock and we expect that the shock displacement rapidly returns to the condition a(t) = 0, since here the homogeneous ODE holds and the solution decays very rapidly. Assume that at the moment of impact, f = 0, exactly half the weight of the bike and rider is absorbed by the shock. For t > ( the front wheel is vertically motionless and in contact with the ground, hence the ODE (1) governs the shock's behavior. At the instant that the bike impacts the ground we have "(0) = 0, since the shock was not compressed in the air. The behavior of the shock for { > 0 can be determined from knowledge of u'(0). A standard physics result shows that an object that falls from a distance h under gravitational acceleration hits the ground with speed v2gh, if air re- sistance is negligible. Therefore, we estimate that the bike hits the ground at a speed of 20 2 -5.42 meter per second, negative because the bike is falling. Thus we take a'(0) = 20. Modeling Exercise 3 Use the initial data w(0) = 0 and w'(0) = -5.42 to find the constants c, and e, in the general solution, and then plot the resulting solution of the initial value problem. Modeling Exercise 4 Compute the maximum compression of the shock that occurs in Modeling Exercise 3 to three significant figures. Suppose the 2bike's front shock has a range of motion of 140 mm before bottoming out (the shock has reached the end of its travel and can no longer compress). Would that be a problem in this scenario? Modeling Exercise 5 (a) Redo the solution for the ODE (1), but change the damping constant from c = 1700 to c = 104, so the system is heavily overdamped. Plot the solution and redo Modeling Exercise 4 with these parameters. What disadvantage might such a value of c have for the rider? (b) Redo the solution for the ODE (1), but change the damping constant from e = 1700 to c = 1200. Show that the system is now underdamped. Plot the solution and redo Modeling Exercise 4. What disadvantage might an underdamped system have for the rider? The goal in this project is to design a front shock absorber that, under the conditions of the examples above: . Has a spring that is as compliant (least stiff) as possible, but yields a shock displacement of no more than 140 mm when the rider rides off a 1.5 meter drop. . Is not excessively overdamped (which makes riding on rugged terrain feel harsh) or underdamped (which makes the bike feel too bouncy.) Modeling Exercise 6 Suppose that the ODE that governs the shock dis- placement y(f) is (as modeled in Example 4.1) my"(t ) + ey (t ) + ky(1) = -mg. (2) where m = 46 kg and g = 9.8. However, let us leave c and k undefined for the moment. These are the parameters in which we are interested. To sim- plify matters, let's start with a shock that is critically damped. What choice "* for c yields a critically damped system for the homogeneous version of (2)? It should depend on k. Write out a general solution to the homogeneous ODE my" +ey' (t) + ky(t) =0. Modeling Exercise 7 Use the method of undetermined coefficients to find a particular solution yo(t) to (2); this solution will depend on k. Then write out a general solution to (2). The general solution should also depend on A.Modeling Exercise 8 As noted previously, a rider who rides off a 1.5 meter drop will hit the ground at a speed of about 5.42 meters per second. Solve (2) with initial conditions y(0) = 0, y'(0) = -5.42, to find the displacement y(t) of 3 the shock. This is a function of t that also involves the indeterminate k. Modeling Exercise 9 Determine, either graphically or analytically, the smallest value & = k* for k that results in the shock compressing no more than -0.14 meters. What is the corresponding value for c*? Modeling Exercise 10 With the values for &* and c* found in Modeling Exercise 4, plot the solution y(f) on the interval 0