Chapter 2: Linear Constant Coefficient Higher Order Equations

Transcription

1 Chapter 2: Linear Constant Coefficient Higher Order Equations The wave equation is a linear partial differential equation, which means that sums of solutions are still solutions, just as for linear ordinary differential equations. So now we can form lots of different solutions out of our various collections of solutions. Usually we will want to find solutions that have particular properties. For example, if our string is tacked down at x = 0 and x = π so that the amplitude of vibration is always 0 at those points, then we want u(0, t) = u(π, t) = 0 for all t. This condition will eliminate many of the different solutions given above (see problem 5). It turns out that all solutions of the wave equation satisfying the condition that the endpoints are tacked down can be written as linear combinations of the solutions we have found, so we will have the general solution to the wave equation subject to this condition. But the situation is more complicated than for ordinary differential equations, because while we have eliminated many of the different solutions with our tacking down condition, we still have infinitely many solutions left so our general solution has infinitely many arbitrary constants. But that is a problem best left to a later course in Fourier series, or better yet partial differential equations. For now, we will settle for noting that the linear constant coefficient ordinary differential equations we have been considering in this chapter often come up in solving linear constant coefficient partial differential equations. Exercises: (1) Find solutions in the form u(x, t) = X(x)T (t) for the heat equation u/ t = κ 2 2 u/ x 2. (2) Find solutions in the form u(x, y) = X(x)Y (y) for Laplace s equation 2 u/ x u/ y 2 = 0. (3) Find solutions in the form u(x, t) = X(x)T (t) for the equation u/ t = 2 u/ x 2 u (heat equation with cooling). (4) Find solutions in the form u(x, y) = X(x)Y (y) for Poisson s equation 2 u/ x u/ y 2 = u. (5) Show that the only solutions of the form u(x, t) = X(x)T (t) to the wave equation that satisfy u(0, t) = u(π, t) = 0 for all t are u(x, t) = (C 1 cos(cnt) + C 2 sin(cnt)) sin(nx) where n is a positive integer. 11 Free Motion Consider a mass on a spring. The forces acting on the mass are gravity with a force mg, where m is the mass and g is the acceleration of gravity (9.8m/sec 2 ), the restoring force of the spring with a force kl, where k is the spring constant, l is how much the spring is 108

2 11 Free Motion stretched and the negative sign denotes the force is pulling toward l = 0, and the damping of the spring (friction) acting with a force cv, where c is the damping constant, v is the velocity and the negative sign denotes that the force acts in opposition to the motion. Now we choose our coordinates for the problem to have x = 0 at the equilibrium length of the spring s, where mg = ks. Note that with this choice, l = x + s. We also choose x > 0 to be when the spring is stretched farther down from equilibrium (this is why we use mg for gravity and not mg). So the total force on the spring is mg kl cv = mg k(s + x) cv = mg ks kx cv = kx cv But by Newton s second law of motion, force = ma where m is mass and a is acceleration. So ma = kx cv Now if we remember that velocity is the derivative of position with respect to time and acceleration is the second derivative of position with respect to time, we have our final equation for free motion of a spring-mass system m d2 x dt 2 + cdx dt + kx = 0 where m is mass, c is the damping constant, k is the spring constant, x is position and t is time. We now solve this equation in general and see what we can deduce about the behavior of such a system. First we assume there is no damping (c = 0). Then the general solution is x(t) = a cos(ωt) + b sin(ωt) = A cos(ω(t + φ)) depending on which way you want to write the formula. Here ω is k/m. This is simple harmonic motion with amplitude A, circular frequency ω and phase shift φ. Next, suppose c > 0. Here there are three different possibilities, depending on whether the discriminant c 2 4km is positive, negative or zero. We look for a solution in the form e rt and obtain (mr 2 + cr + k)e rt = 0 109

3 Chapter 2: Linear Constant Coefficient Higher Order Equations from which we find r = c ± c 2 4km 2m Now if the discriminant is positive, this equation has two distinct real roots, call them r 1 and r 2. The general solution is then c 1 e r 1t + c 2 e r 2t. We note that r 1 and r 2 will both be negative, so the solution will decay to 0 monotonically. In this case the system is said to be overdamped (see figure 1). If the discriminant is negative, then the roots will be complex conjugates. The real part of the roots will be c/2m which is negative so the solution will still decay to 0, but now it will oscillate while doing so, since the imaginary part of the roots will give rise to a factor of cos(ω(t+p)) in the solution. In this case the system is said to be underdamped (see figure 2). Finally, it is possible for the discriminant to be zero. Here we have a double root at c/2m and the general solution will be c 1 e (c/2m)t + c 2 te (c/2m)t This solution will converge monotonically to 0. In this case the system is said to be critically damped (see figure 3). Note that there is no obvious distinction between the graph of an overdamped and a critically damped spring. Exercises: (1) A mass of 1kg is attached to an undamped spring, which causes the spring to stretch 10cm. The spring is then pulled down an additional 20cm and released. What is the equation of the resulting motion? (2) A mass of 5kg is attached to an undamped spring, which causes the spring to stretch 5cm. The mass is then given an initial velocity of 12m/sec (in the downward direction). What is the equation of the resulting motion? (3) A mass of 100g is attached to an undamped spring, which causes the spring to stretch 5cm. The mass is then pulled down an additional 5cm and then released. What is the equation of the resulting motion? (4) A mass of 1kg is attached to an undamped spring, which causes the spring to stretch.1m. The mass is then pulled down an additional.1m and then released. What is the equation of the resulting motion? (5) What is your answer to question 4 if the spring is on the moon instead of the earth? (The acceleration of gravity for the moon is 1/6 that for the earth). (6) A spring mass system has a mass of 1kg, a spring constant of 6kg/sec 2, and a damping constant of 1kg/sec. If the mass is pulled 1m from equilibrium and released, what is the equation of the resulting motion? (7) Suppose an additional damping mechanism is added to the system of problem 6 so the damping constant becomes 5kg/sec. What would the equation of the resulting motion be 110

4 11 Free Motion Figure 1 Figure 2 Figure 3 111

5 Chapter 2: Linear Constant Coefficient Higher Order Equations in this case? (8) A spring mass system has a mass of 50g, a spring constant of 60g/sec 2, and a damping constant of 10g/sec. If the mass is pulled 10cm from equilibrium and released, what is the equation of the resulting motion? (9) Suppose an additional damping mechanism is added to the system of problem 8 so the damping constant becomes 40g/sec. What would the equation of the resulting motion be in this case? (10) Consider an underdamped spring mass system with a mass m > 0, a damping constant 0 < c < 2 km, and a spring constant k > 0. What is the general solution of the corresponding differential equation? (11) The motion in problem 10 is not periodic because the amplitude varies. Show that it is quasiperiodic, that is, the time between successive maxima is constant. What is the quasiperiod? (12) Show that the ratio of successive maxima in problem 10 is constant. What is this ratio? (13) Suppose a mass of 40g is attached to a spring, which causes the spring to stretch 6cm. The spring is then set in motion and the quasiperiod is observed to be 30 seconds. What is the damping constant of the spring? (14) Suppose a mass of 1000kg is attached to a spring, which causes the spring to stretch 10cm. The spring is then set in motion and the quasiperiod is observed to be 10 seconds. What is the damping constant of the spring? Get into Matlab and give the command lab7. This will create a window where you can graph the solutions of second order constant coefficient linear differential equations. You can change values in the white fields. The mass and spring values must be positive constants, and the damping value must be a non-negative constant. When you change a value, two buttons labeled Okay and Cancel will appear in the lower right corner. If you press Okay, the graphs will be updated. If you press Cancel, the white fields will be reset to their previous values. This way you can change several values at once and update the graphs only when all the changes are typed in. While you are editing values, the graph will appear on a gray background to remind you that the displayed graph doesn t correspond to the currently displayed values. The background will go back to black once you press Okay or Cancel and the displayed graph does correspond to the displayed values again. Also you can choose to display either one set of initial conditions or two sets of initial conditions at once using the Initial Values menu at the top of the window. 112

6 12 Forced Motion For problems 15 through 20 you need to find equations and initial values that give graphs similar to the given pictures. (You probably won t be able to get graphs that look exactly like the given pictures, but do the best you can.) In addition to listing your equations and initial values, you need to turn in a brief discussion (one or two sentences) of what is being illustrated in each picture. (15) (16) (17) (18) (19) (20) 12 Forced Motion Discussion: We now consider a spring-mass system subject to an external force. The most interesting forces are periodic forces, which we will assume take the form F (t) = F 0 cos(ωt) where F is force, t is time, F 0 is amplitude and ω is the circular frequency. We start with the undamped case (c = 0). We have solved the homogeneous problem in the previous 113

7 Chapter 2: Linear Constant Coefficient Higher Order Equations section, obtaining A cos(ω 0 (t + φ)) where ω 0 = k/m is called the natural frequency of the spring-mass system. We will now consider the particular solution of m d2 x dt 2 + kx = F 0 cos(ωt) Since F 0 cos(ωt) = R[F 0 e iωt ], we will try to find the particular solution for the complex problem and then take the real part. So we guess So we get z = ae iωt z = iωae iωt z = ω 2 ae iωt mω 2 ae iωt + kae iωt = F 0 e iωt a = F 0 mω 2 + k = F 0 m(ω 2 0 ω2 ) and taking the real part of ae iωt yields a particular solution of x(t) = F 0 m(ω 2 0 ω2 ) cos(ωt) Note that the particular solution has the same frequency as the forcing function but that the amplitude is divided by a factor m(ω 2 0 ω 2 ). We would expect to see that mass would be inversely proportional to the displacement generated by a fixed force, but it may be something of a surprise that the frequency plays such a large role. The underlying physical intuition is that the spring wants to oscillate at its natural frequency, and the closer the forcing function is to the natural frequency, the more the forcing function and the spring will work together and not in opposition, hence the larger amplitude of the response. The general solution is then x(t) = F 0 m(ω 2 0 ω2 ) cos(ωt) + A cos(ω 0(t + φ)) If we plot out the solution curves, we will see the phenomenon of beats arising (see figure 4). This is most easily seen if ω and ω 0 are close. Then the amplitude of the solution will steadily increase as the forcing function pours ever greater amounts of energy into the system. However, if the forcing function is not exactly in sync with the natural frequency of the system (that is ω ω 0 ) then eventually the forcing function will become out of 114

8 12 Forced Motion Figure 4 phase with the natural frequency. Then the force applied to the system will reduce the amplitude and the beat will die off. At the end of the beat, the forcing function will have worked its way back in sync with the natural frequency and the pattern will start all over again. Of course, if ω = ω 0, then we have divided by zero and our solution is not valid for this case. Here, we get a particular solution of the form x(t) = F 0 2mω 0 t sin(ω 0 t) You may check that this is indeed a particular solution. amplitude grows linearly with t. Note that in this case, the Since the forcing function and the natural frequency are exactly in sync, the forcing function always acts to add energy to the system and the amplitude grows without bound, at least mathematically. In real life, the system will tear itself apart eventually as the amplitude gets too large. This is the phenomenon of resonance (see figure 5). Next we consider what happens when we include damping (c > 0). We will consider just the underdamped case (the overdamped is similar). The homogeneous solution is x(t) = Ae (c/2m)t cos(ω 1 (t + φ)) where ω 1 = ω0 2 (c/2m)2 and ω 0 = k/m is the natural frequency. Note that if c is small then the spring-mass system oscillates with circular frequency ω 1 very close to the natural frequency ω

9 Chapter 2: Linear Constant Coefficient Higher Order Equations Figure 5 Now when we find the particular solution, we again use F 0 cos(ωt) = R[F 0 e iωt ] and solve for a particular solution to the complex problem and take the real part. We guess and so we find z = ae iωt z = iωae iωt z = ω 2 ae iωt mω 2 ae iωt + ciωae iωt + kae iωt = F 0 e iωt ( (k mω 2 ) + icω ) ae iωt = F 0 e iωt F 0 a = (k mω 2 ) + icω F 0 a = m(ω0 2 ω2 ) + icω Here there is no danger of dividing by zero. Even if ω 0 = ω, c > 0 so the denominator is non zero. Now let R = m 2 (ω 2 0 ω2 ) 2 + c 2 ω 2 be the modulus of m(ω 2 0 ω 2 ) + icω and let Φ = arctan(cω/[m(ω 2 0 ω 2 )]) be the argument of m(ω 2 0 ω 2 ) + icω. Then the real particular solution will be R The general solution is [ F0 eiωt ReiΦ ] [ ] F0 = R R ei(ωt Φ) = F 0 cos(ωt Φ) R F 0 R cos(ωt Φ) + e ct/2m A cos(ω 1 t + φ) 116

10 12 Forced Motion Figure 6 The first term (the particular solution) is called the steady state solution and the second term (the homogeneous solution) is called the transient solution. Note that the amplitude of the transient term decays exponentially, so it will become unimportant as t grows large, hence the name transient. The steady state term has a constant amplitude, hence the name steady state (see figure 6). With positive damping, R will never be zero so the solution will not have amplitude that increases without limit. On the other hand, if ω 0 = ω then R = c and if c is small, it is still possible for the response to be sufficiently large to break the spring-mass system. Exercises: (1) A mass of 2kg is attached to an undamped spring, which causes the spring to stretch 15cm. The mass is subject to an external force of cos(2t) Newtons. What is the equation of the resulting motion? (2) A mass of 3kg is attached to an undamped spring, which causes the spring to stretch 10cm. The mass is subject to an external force of 5 cos(2πt) Newtons. What is the equation of the resulting motion? (3) A mass of 50g is attached to an undamped spring, which causes the spring to stretch 5cm. The mass is subject to an external force of 10 cos(10t) dynes (a dyne is a g cm/sec 2 ). What is the equation of the resulting motion? (4) A mass of 10g is attached to an undamped spring, which causes the spring to stretch 2cm. The mass is subject to an external force of 5 sin(2πt) dynes (a dyne is a g cm/sec 2 ). What is the equation of the resulting motion? (5) A mass of 1kg is attached to an undamped spring, which causes the spring to stretch 12cm. The mass is subject to an external force of cos(ωt) Newtons. For what value of ω 117

11 Chapter 2: Linear Constant Coefficient Higher Order Equations does resonance occur? (6) A mass of 10g is attached to an undamped spring, which causes the spring to stretch 12cm. The mass is subject to an external force of cos(ωt) Newtons. For what value of ω does resonance occur? (7) A mass of 1kg is attached to a spring, which causes the spring to stretch 40cm. The damping constant of the spring is 5kg/sec. The mass is subject to an external force of 30 cos(100t) Newtons. What is the equation of the resulting motion? (8) A mass of 30g is attached to a spring, which causes the spring to stretch 20cm. The damping constant of the spring is 1kg/sec. The mass is subject to an external force of 4 cos(5t) Newtons. What is the equation of the resulting motion? (9) A mass of 120g is attached to a spring, which causes the spring to stretch 12cm. The damping constant of the spring is 1kg/sec. The mass is subject to a constant external force of 6 Newtons. What is the equation of the resulting motion? (10) A mass of 50g is attached to a spring, which causes the spring to stretch 10cm. The damping constant of the spring is 500g/sec. The mass is subject to a constant external force of 10 cos(2πt) dynes (a dyne is a g cm/sec 2 ). What is the equation of the resulting motion? Electric circuits are analyzed using exactly the same equations as spring mass systems, with inductance, L, measured in henrys, playing the role of mass, resistance, R, measured in ohms, playing the role of the damping constant, and elastance playing the role of the spring constant. Elastance is defined as 1/C, where C is the capacitance, which is measured in farads (though typical capacitors may have their capacitance stated in microfarads = 10 6 farads). The role of the external force is played by the impressed voltage, E(t), measure in volts. Charge, Q, measured in coulombs plays the role of position (x) and current, I, measured in amperes (which are coulombs/sec) plays the role of velocity. (11) Determine the steady state current in a circuit consisting of a coil, with inductance 1 henry, a resistor, with resistance 200 ohms, a capacitor, with capacitance 10 microfarads, and an impressed voltage of 170 cos(120πt) (which corresponds to the output of a standard electric outlet in this country). (12) Determine the steady state current in a circuit consisting of a coil, with inductance 0.5 henry, a resistor, with resistance 100 ohms, a capacitor, with capacitance 2 microfarads, and an impressed voltage of 170 cos(120πt). (13) Determine the steady state current in a circuit consisting of a coil, with inductance 1 henry, a resistor, with resistance 200 ohms, a capacitor, with capacitance 4 microfarads, 118

12 12 Forced Motion and an impressed voltage of 340 cos(100πt). (14) Determine the steady state current in a circuit consisting of a coil, with inductance 0.1 henry, a resistor, with resistance 80 ohms, a capacitor, with capacitance 1 microfarads, and an impressed voltage of 340 cos(100πt). (15) Determine the steady state current in a circuit consisting of a coil, with inductance 0.5 henry, a resistor, with resistance 150 ohms, a capacitor, with capacitance 5 microfarads, and an impressed voltage of 6volts (this is direct current). (16) Determine the steady state current in a circuit consisting of a coil, with inductance 0.2 henry, a resistor, with resistance 100 ohms, a capacitor, with capacitance 2 microfarads, and an impressed voltage of 12 volts. (17) Suppose a circuit offers no resistance (superconductance). If the circuit has a coil with inductance 5 henrys, what capacitance would you choose for a capacitor to make the circuit resonate at a frequency of 60Hz? Note that Hz stands for cycles/sec, and that you must multiply by 2π to get the circular frequency (radians/sec) which is used in the formulas in the text. (18) Suppose a circuit has inductance.1 henry, resistance 80 ohms, capacitance 4 microfarads and an impressed voltage of 170 cos(120πt). What is the amplitude of the steady state current? The impedance of the circuit is defined as the ratio of the amplitude of the impressed voltage (170 in this case) to the amplitude of the steady state current. What is the impedance of the circuit? (19) Suppose a circuit has inductance.2 henry, resistance 100 ohms, capacitance 2 microfarads and an impressed voltage of 340 cos(100πt). What is the amplitude of the steady state current? The impedance of the circuit is defined as the ratio of the amplitude of the impressed voltage (340 in this case) to the amplitude of the steady state current. What is the impedance of the circuit? (20) Suppose a circuit has inductance L, resistance R, capacitance C and an impressed voltage of E 0 cos(ωt). What is the amplitude of the steady state current? The impedance of the circuit is defined as the ratio of the amplitude of the impressed voltage (E 0 ) to the amplitude of the steady state current. What is the impedance of the circuit? Get into Matlab and give the command lab8. This will create a window where you can graph the solutions of second order constant coefficient linear differential equations. You can change values in the white fields. The mass and spring values must be positive constants, the damping value must be a non-negative constant and the forcing function must be a function of t. When you change a value, two buttons labelled Okay and Cancel will appear in the lower right corner. If you press Okay, the graphs will be updated. If 119

13 Chapter 2: Linear Constant Coefficient Higher Order Equations you press Cancel, the white fields will be reset to their previous values. This way you can change several values at once and update the graphs only when all the changes are typed in. While you are editing values, the graph will appear on a gray background to remind you that the displayed graph doesn t correspond to the currently displayed values. The background will go back to black once you press Okay or Cancel and the displayed graph does correspond to the displayed values again. Also you can choose to display either one set of initial conditions or two sets of initial conditions at once using the Initial Values menu at the top of the window. For problems 21 through 25 you need to find equations and initial values that give graphs similar to the given pictures. In addition to listing your equations and initial values, you need to turn in a brief discussion (one or two sentences) of the what is being illustrated in each picture. (21) (22) (23) (24) (25) 120

14 13 Energy of a Spring Mass System 13 Energy of a Spring Mass System Discussion: You probably studied spring mass systems in your freshman physics course. In the usual analysis given in physics courses, the concept of energy plays an important role. I shall now prove that energy is the only conserved quantity in a spring mass system. Suppose we have a mass m attached to an undamped spring with spring constant k. Let x denote the position of the mass with x = 0 being the equilibrium position. Then, as any text shows, the motion of the mass satisfies the differential equation (1) m d2 x + kx = 0. dt2 Following the physicists, we let v = dx/dt. The total energy of this system is E(x, v) = 1 2 mv kx2 where 1 2 mv2 is the kinetic energy and 1 2 kx2 is the potential energy of the spring (see any freshman physics text). An easy calculation then shows de dt = d ( 1 dt 2 mv2 + 1 ) 2 kx2 = mv dv dt + kxdx dt = mv d2 x dt 2 + kxv ( ) = v m d2 x dt 2 + kx = 0. This shows that energy is conserved in the undamped system. We now show that energy is the only conserved quantity. We must be a little more careful and precise now. Obviously, Φ(E) is conserved for any function Φ. function G(x, v) which is conserved then we can write G = Φ(E). What we shall show is that given any smooth Theorem 1. Let G(x, v) be a continuously differentiable function such that G(x, v) is constant whenever x solves (1) and v = dx/dt. Then the level curves of G are the same as the level curves of E. Proof: Since G(x(t), v(t)) is constant we have dg/dt = 0. functions of several variables we rewrite this as G dx x dt + G dv v dt = 0. Using the chain rule for 121

15 Chapter 2: Linear Constant Coefficient Higher Order Equations Using (1) and multiplying through by m we can write this as mv G x kx G v = 0. This is a linear first order partial differential equation. We will now solve this by applying some standard results from multivariable calculus. The geometric methods are a special case of the standard method of characteristics for solving first order partial differential equations which you can learn about in later courses. We rewrite our equation as (mv, kx) G = 0. Since level curves are orthogonal to the gradient of a function, the vector (mv, kx) must be tangent to the level curves of G. Hence the slopes of the level curves of G must be given by kx/mv. So the level curves of G satisfy the separable first order ordinary differential equation dv dx = kx mv, whose general solution is 1 2 mv kx2 = K. But this says that the level curves of G are exactly the level curves of E, which was to be proved. This implies the claim that G(x, v) = Φ(E(x, v)). Since the level curves of E and G are the same, then if you know the value of E, even if you don t know the values of x and v, then you know which level curve of E you are on and hence which level curve of G you are on and hence you can determine the value of G just from a knowledge of the value of E, so G must be some function of E. We now go over a quick example of using energy to analyze a spring mass system. EXAMPLE: A mass of 1kg is attached to an undamped spring which stretches 0.5m. The mass is pulled down an additional.1m and released with a downward velocity of 2m/sec. What is the amplitude of the resulting motion? A mathematical way to solve the problem is to find the function for position versus time by solving the initial value problem x x = 0, x(0) =.1, x (0) = 2 and observing what the amplitude is. But the physical approach using energy is much easier. Energy = E = 1/2mv 2 + 1/2kx 2 = 1/ / = =

16 13 Energy of a Spring Mass System But at the maximum extension, the motion is stopped (mathematically this says x is at a maximum when v = x = 0) and all the energy is potential energy, 1/2kx 2. So we get Potential Energy = 1/2kx 2 set = / x 2 = x = 2.098/9.8.5m to one significant figure (which is all the accuracy we have in the data). You should now be able to sketch the motion fairly easily by computing the natural circular frequency ω 0 = k/m = to one significant figure, and drawing a cosine wave with this circular frequency using the amplitude we computed above and the given initial values (see figure 7). Figure 7 We now turn our attention to the damped system (2) m d2 x dt 2 + cdx dt + kx = 0 where c > 0. In your physics class this system probably received much less attention. This is because it is not only more realistic, it is also much more difficult to analyze. Using the same calculation as given for the undamped system shows de/dt = cv 2 0 in this case. In this case, there are no non trivial continuous conserved quantities. Without conserved quantities, there is no way to avoid the sort of calculations we have used in the previous section. 123

17 Chapter 2: Linear Constant Coefficient Higher Order Equations Theorem 2. Let H(x, v) be a continuous function such that H(x(t), v(t)) is constant whenever x(t) solves (2) and v = dx/dt. Then H(x, v) is constant for all (x, v). Proof: The general solution to (2) is ( ) 4km c x(t) = e ct/2m 2 A cos t + p 2m where A and p are constants. Differentiating we find [ ( ) v(t) = e ct/2m c ( )] 4km c 2m A cos 2 4km c 2 4km c 2 t + p A sin t + p. 2m 2m 2m So lim t (x(t), v(t)) = 0. Since H(x, v) is continuous, lim t H(x(t), v(t)) = H(0, 0). But since H(x(t), v(t)) is constant, it follows that H(x, v) = H(0, 0) for all (x, v). We can use techniques like those in the proof of theorem 1 to find conserved quantities which are discontinuous at (x, v) = (0, 0). Because of the discontinuity however, these quantities are of no real physical interest. Exercises: (1) A mass of 3kg is attached to an undamped spring, which causes the spring to stretch 15cm. The mass is then pulled down 10cm and released. Sketch the resulting motion. (2) A mass of 140g is attached to an undamped spring, which causes the spring to stretch 25cm. The mass is given an initial velocity of 2m/sec in the downward direction. Sketch the resulting motion. (3) A mass of 60g is attached to an undamped spring, which causes the spring to stretch 5cm. What velocity must the mass be given to pull the spring an additional 15cm? (4) A mass of 1kg is attached to a spring, which causes the spring to stretch 10cm. The spring has a damping constant of 1kg/sec. The mass is pulled down 25cm and released. Write the energy of the system as a function of time. (5) A mass of 1kg is attached to a spring, which causes the spring to stretch 10cm. The spring has a damping constant of 2kg/sec. The mass is pulled down 25cm and released. Write the energy of the system as a function of time. (6) A mass of 1kg is attached to a spring, which causes the spring to stretch 10cm. The spring has a damping constant of 3kg/sec. The mass is pulled down 25cm and released. Write the energy of the system as a function of time. 124

18 14 Systems of Equations (7) A mass of 1kg is attached to a spring, which causes the spring to stretch 5cm. The spring has a damping constant of 3kg/sec. The mass is subjected to an external force of 10 cos(t) Newtons. Write the energy of the system as a function of time. (8) A mass of 25g is attached to an undamped spring which causes the spring to stretch 1cm. The mass is subjected to an external force of 5 dynes (grams per second squared). Write the energy of the system as a function of time. (9) A mass of 25g is attached to an undamped spring which causes the spring to stretch 1cm. The mass is subjected to an external force of 5 cos(20t) dynes (grams per second squared). Write the energy of the system as a function of time. (10) A mass of 1kg is attached to an undamped spring which causes the spring to stretch 5cm. The mass is subject to an external force of 2 cos(14t) Newtons. Write the energy of the system as a function of time. 14 Systems of Equations The final topic we will cover this semester is systems of equations. For example, suppose x(t) and y(t) are both functions of t and satisfy the system of equations dx = 3x + 5y dt dy dt = 3x + y x(0) = 1 y(0) = 2. Systems of first order equations arise in population dynamics when you deal with two interacting populations, traditionally foxes and rabbits. Systems of second order equations arise in electrical systems when you have a circuit with more than one loop and in springmass systems when you have several different springs connected together. There are a variety of techniques for solving systems of linear differential equations such as the example above. One way is to use some algebraic manipulations to convert the first order linear system to a second order linear equation for x which you can solve by the techniques of chapter 2. EXAMPLE: Find the general solution to dx = 3x + 5y dt dy dt = 3x + y 125