HOMEWORK 4: MATH 265: SOLUTIONS. y p = cos(ω 0t) 9 ω 2 0

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1 HOMEWORK 4: MATH 265: SOLUTIONS. Find the solution to the initial value problems y + 9y = cos(ωt) with y(0) = 0, y (0) = 0 (account for all ω > 0). Draw a plot of the solution when ω = and when ω = 3. Solution: First solve the homogeneous problem. Let y = e rt. Then, the characteristic polynomial is r = 0, so that r = ±3i. Thus, y = sin(3t) and y = cos(3t) are solutions to the homogeneous problem. Find the particular solution when ω 0 3. The particular solution is easily found to be The general solution, when ω 0 3, is y p = cos(ω 0t) 9 ω 2 0 y = c cos(3t) + c 2 sin(3t) + cos(ω 0t) 9 ω 2 0 Satisfying the initial conditions y(0) = y (0) = 0, we get c = 0, c 2 = /(9 ω0). 2 Thus, the solution to the IVP is y = (9 ω0 2) [cos(ω 0t) cos(3t)], when ω 0 3. (2) When ω 0 = 3 we have resonance and y p = (t/6)sin(3t). The general solution is thus, y = c cos(3t) + c 2 sin(3t) + t 6 sin(3t). Satisfying the initial conditions we get c = c 2 = 0. Hence, the solution to the initial value problem is.. y = t 6 sin(3t), when ω 0 = 3. (3) In Fig. a and Fig. b we plot y versus t when ω 0 = and when ω 0 = 3, respectively. Notice that when ω 0 = 3 we get an unbounded oscillation. ¼ ¾ ¾ ¼ Ý µ Ý µ ¼ ¾ ½ ¾ ¼ ½ (a) y(t) when ω 0 = 3 (b) y(t) when ω 0 = (resonance) Figure : Left figure: Plot of y(t) = 8 (cos(t) cos(3t)) when ω 0 = 3. Right figure: Plot of y(t) = t 6 sin(3t) when ω 0 =. This is the resonance case.

2 2. Charging of a capacitor: Consider the RLC circuit, where the resistor, inductor and capacitor are in series with an external D.C. voltage source which is active for 0 < t < T but is then disconnected from the circuit at t = T. The fixed time T is called the switching time. Then, the charge Q(t) on the capacitor satisfies LQ + RQ + C Q = { V0 for 0 t < T, 0 for t T, where V 0 is a constant. Assume that Q(0) = Q (0) = 0, and that R 2 > 4L/C. Find the solution Q = Q(t) and sketch two plots of Q(t), one for large T and one for small T. Explain the result on physical terms. Solution: Let Q = e rt in the homogeneous problem to get Lr 2 + Rr + /C = 0, so that r ± = R 2L ± (R2 4L/C) 2L The roots are real and negative since R 2 > 4L/C. The particular solution is Q = CV 0 for 0 < t < T. Hence, we have { c e Q(t) = r+t + c 2 e r t + CV 0 for 0 t < T, d e r+(t T) + d 2 e r (t T) () for t > T. There are four constants to determine: c, c 2, d and d 2. Satisfying the initial conditions we get two equations c + c 2 + CV 0 = 0 and r + c + r c 2 = 0. Thus, c 2 = r +CV 0 (r + r ), c = r CV 0 (r + r ), where r (R2 4L/C) + r =. (2) L Now we satisfy the conditions that Q and Q are continuous at t = T. This yields two equations for d and d 2. We get d + d 2 = c e r+t + c 2 e r T + CV 0, r + d + r d 2 = c r + e r+t + c 2 r e r T. We can solve for d and d 2 to get d = c e r+t. CV 0r (r + r ), d 2 = c 2 e r T + CV 0r + (r + r ). (3) Substituting (2) and (3) into () gives Q = Q(t). Notice that when T is large (i.e. r + T and r T ), then we can neglect the exponential terms in (3). In addition, the exponential terms in the formula for Q(t), defined on the interval 0 t < T, are insignificant when t T. Thus, the capacitor has had time to be almost fully charged and Q CV 0 when t T. The plots of the solutions are shown below in Fig. 2 for T small (the capacitor is not fully charged) and for T large when the capacitor is fully charged. In both cases when t > T, the charge on the capacitor leaks out as t since the circuit has resistance. ¼ Õ µ ¼ ¼ ¼ ¾ ½ ½ ¾ ¾ Figure 2: Plot of q(t) versus t for T = 5 (heavy solid curve) and T = 20 (solid curve). The parameters are C = V 0 = L = and R = 20. 2

3 3. Solve y + 25y = cos(5.6t) with y(0) = 0 and y (0) = 0. Give an accurate plot of the solution. What do you observe? Solution: For the homogeneous problem let y = e rt. Then, r = 0 so that r = ±5i. The solution to the homogeneous problem is y = sin(5t) or y = cos(5t). Now calculate the particular solution. Let y p solve y p + 25y p = cosωt, ω 5.6 The particular solution is found (by substituting y p = Acosωt) to be The general solution is y p = cos(ωt) 25 ω 2. y = c cos(5t) + c 2 sin(5t) + cos(ωt) 25 ω 2. Satisfying the initial conditions y(0) = y (0) = 0, we get c 2 = 0, c = /(25 ω 2 ). Thus, using a trig identity we can write y = (25 ω 2 [cos(ωt) cos(5t)] = ) 3.8 sin(.3t)sin(5.3t). The solution, which exhibits the phenomenon of beats as described in class, is plotted below in Fig. 3. ¼ ¼ ¾ Ý µ ¼ ¼ ¾ ¼ ½ ½ ¾ Figure 3: Plot of y(t) = sin(0.3t)sin(5.3t) versus t. 3

4 4. Consider y + py + y = F 0 sin(ωt), where p > 0 and F 0 are constants. Plot the amplitude of the steadystate response as a function of ω on the interval ω > 0. Draw these plots for several representative values of p. Where is the amplitude a maximum as a function of ω? Interpret the results on physical grounds with the idea that p represents a damping term. Solution: A simple calculation (using y p = Acosωt+B sin ωt and the trig formula cos(a±b) = cosacosb sin Asin B) yields the particular solution where R(ω) and φ(ω) satisfy R(ω) = y p = R(ω)cos(ωt φ(ω)), F 0 [( ω 2 ) 2 + p 2 ω 2 ], tan(φ) = ( ω2 ), ( ) /2 pω with cos(φ) < 0. The amplitude of the steady-state response is R(ω). It has a maximum, on the interval ω 0, when the denominator in ( ) has a minimum on this interval. Using a little calculus, we find that this occurs at ω = ω c, where ω c = p2 2, 0 < p < 2 ; ω c = 0, p 2. Thus, if the damping coefficient p is not too large (i.e. p < 2), then the system has its largest response at a frequency ω c 0. The maximum amplitude R(ω c ) increases as p decreases and becomes unbounded as p 0. The frequency ω c tends to the resonant frequency of as p 0. When F 0 =, we plot R(ω) versus ω below in Fig. 4 for p 0, for p =, p = 2 and p > 2. ¼ Ê µ ¾ ¼ ¾ ¼ ¼ Figure 4: Plot of R(ω) in (*) versus ω for p = 0.2 (heavy solid curve), p =.0 (solid curve), p = 2 (dotted curve), and p = 2 (widely spaced dots). 4

5 5. (Tuning a circuit) Consider an RLC circuit in series with an A.C. voltage source given by V (t) = cost 4/5cos(5t). Suppose that R = 0. Ohms, L = Henry but that we are capable of adjusting the capacitance C of the capacitor. The current in the circuit satisfies (i) Calculate the steady-state solution. I + 0.I + I = sin t + 4 sin(5t). C (ii) What are the critical values of C for which the current I will (roughly) be periodic with a frequency of either or 5? Interpret this result in terms of the tuning of a circuit. (iii) Give a sketch of the steady-state solution when C =, C = /25 and C = /8. Solution: (i) Let I ω (t) be the steady state response for Then, by linearity, the steady-state response for ( ) is I ω + 0.I ω + C I ω = sin(ωt). (2) I(t) = I (t) + 4I 5 (t). (3) Notice that the solution to the homogeneous problem, (i.e. the transient response) will die out as t since the circuit has resistance. Thus, as t, we will only observe the steady-state response. Now we calculate the particular solution for (2). Consider Ĩ ω + 0.Ĩ ω + C Ĩω = sin ωt. Let Ĩω = Acosωt + B sin ωt. Substituting, we can determine (after a little algebra!) I ω = sin(ωt + φ(ω)) (ω), (ω) [ (ω 2 C ) ω 2] /2, tan(φ(ω)) 0.ω (ω 2 C ), where cos(φ) < 0. Hence, from (3), the steady-state response is I(t) = sin(t + φ()) () sin(5t + φ(5)) + 4. (4) (5) (ii) From the formula above the response which oscillates with frequency will be amplified when C =. We obtain a very large amplification of this term since the damping coefficient given by the resitance is very small. For this value of C, the response which oscillates at a frequency of 5 will have a very small amplitude and hence will not be observed in a significant way. Alternatively, the term which oscillates with frequency 5 will be amplified when C /25. For this value of C, the response which oscillates at a frequency of will have a very small amplitude and hence will not be observed in a significant way. Thus, by varying C the circuit can be tuned to respond to either of the two frequencies. When C = /8, the circuit does not respond to either of the two frequencies of and 5 since both of these oscillations can be seen from (4) to have a very small amplitude. (iii) The plots are shown below for three values of C. 5

6 ¼ ½¼ ½ Á µ Á µ ¼ ½ ½ ¼ ½¼ ½ (a) I(t) when C = and C = /25 (b) y(t) when C = /8 Figure 5: Left figure: Plot of I(t) when C = (heavy solid curve) and when C = /25 (solid curve). Notice that these values of C choose one of the frequencies. Right figure: Plot of I(t) when C = /8. In this case, the current is the superposition of two distinct frequencies and the current has a very small amplitude. 6

1. (10 points) Find the general solution to the following second-order differential equation:

Math 307A, Winter 014 Midterm Solutions Page 1 of 8 1. (10 points) Find the general solution to the following second-order differential equation: 4y 1y + 9y = 9t. To find the general solution to this nonhomogeneous