Please improve my math paper

Anonymous
timer Asked: Dec 12th, 2018

Question description

Please improve my math paper, I paper has some miss, I will upload the example paper, Please look the example paper and improve my paper

Sample Project: Spring-Mass via Systems of ODEs and Phase Portraits MATH527 Fall 2018: Sample Produced by Instructor, John McClain 1. Introduction I study a damped spring-mass system as a single second-order equation and as a system of first-order equations. I discuss phase portraits and trajectories in the case of underdamped and critically damped unforced systems and in the case of a forced, underdamped system. 2. Single Equation 2.1. Derivation and Meaning of Parameters. Let us assume that there are at most three forces acting on the bob: the force that is due to gravity, the force produced by the spring, and possibly a drag force. The force that is due to gravity is given by mass × gravity (mg), and the force produced by the spring is proportional to the amount (length) that the spring has been stretched or compressed (k∆L) from its natural length (this is called Hooke’s law ). The constant k in the spring force is called the spring constant, and it has units dyne/cm or Newton/m. The drag or damping force is proportional to the velocity of the mass, but opposes the motion. The proportionality constant, c, has units of N · s/m. According to Newton’s second law of motion, the equation of motion of the bob is mass × acceleration of the bob = sum of all forces acting on the bob (1) md00 = k∆L − mg = k(∆l − d) − mg − cd0 + fext (t) = −kd − cd0 + fext (t) This gives the differential equation: md00 + cd0 + kd = fext (t). 2.2. Meaning and Relevance of Homogeneous vs Non-Homogeneous. If there is an external force fext (t), the equation is non-homogeneous. If there is no external force, so that fext (t) = 0, the equation is homogenous. The external force models a time-dependent force on the bob, like a motor attached to it. 2.3. General Solution. I picked the following values of the parameters to study: m = 1 kg, c = 6 N · s/m, and k = 9 N/m. I picked an external of fext = 12 cos 3t N. The general solution is: d(t) = c1 e−3t + c2 te−3t + 2 cos 3t. 2.4. IVP and a Paricular Solution. I picked the initial condition that the spring is pushed at 10 m/s upward from 2, m above equilibrium. So, d(0) = 2 m. and d0 (0) = 10. The particular solution is: y(t) = 10te−3t + 2 cos 3t. 2.5. Behavior of This Solution. The bob in this case moves upward at first to its highest point of about 2.8 m above equilibrium, then back downward to below equilibrium. It moves back up to just slightly above 2 m above equilibrium. Note that 2 m is the amplitude of the forcing. After only one cycle of the bob moving up, down, and then back up, its motion appears to be completely determined by the forcing. The effect of the initial condition has decayed to a low level, leaving a motion that is dominated by oscillations in sync with the sinusoidal external force. 1 2 3. System of ODEs 3.1. Derivation of the System of Equations Model and Meaning of Parameters. To get a system of equations from the single, second-order differential equation, we define y1 (t) to be the displacement from equilibrium. Then, we define y2 (t) to be the velocity, the derivative of displacement. Then, the single, second-order equation, md00 + cd0 + kd = fext (t), becomes the system of equations: y10 = y2 y20 = − k c fext (t) y1 − y2 + m m m or as a matrix vector system: 0 y = 0 1 k −m c −m ! y+ where y = ! 0 , fext (t) m y1 y2 ! = d ! . v The parameters here have the same meanings as for the single equation. They now show up as ratios only. 3.2. Meaning and Relevance of Homogeneous vs Non-Homogeneous. Again, the inhomogeneity models an external force on the bob, like that supplied by a AC motor producing sinusoidally-varying forces on the bob. The inhomogeneity for the system must have the form ! 0 f (t) = fext (t) m because the force only appears in the equation for y20 which comes from Newton’s 2nd Law. If there is no external force, the system is homogeneous. 3.3. Homogeneous Systems. I pick two sets of parameters. One {m = 1 kg, c = 1 N ·s/m, k = 17/4 N/m} leads to complex eigenvalues, while the other {m = 4 kg, c = 12 N · s/m, k = 9 N/m} leads to a repeated eigenvalue. (The second set of parameters is the same as I used in my singl equation.) fext (t) = 0 for all t, so the system is homogeneous. 3.3.1. Complex Eigenvalues. With the parameters {m = 1 kg, c = 1 N · s/m, k = 17 4 N/m}, the system can be written: 0 1 −17/4 −1 0 y = ! y. These parameters are realistic, representing a very large strong spring, with a sizeable bob on it. The damping is fairly strong, but the system is still underdamped. The general solution to this system can be written: y = c1 e−t/2 or cos 2t − 21 cos 2t − 2sin2t ! + c2 e−t/2 sin 2t − 21 sin 2t + 2 cos 2t ! 3 Figure 1. Phase Portrait for Damped System, {m = 1 kg, c = 1 N · s/m, k = y=e (− c21 + 2c2 ) cos 2t − (2c1 + N/m} ! c1 cos 2t + c2 sin 2t −t/2 17 4 c2 2 ) sin 2t 3.3.2. Phase Portrait. Fig. 1 shows a phase portrait for this system. All solutions spiral toward the origin because the two eigenvalues are complex with negative imaginary parts. 3.3.3. Initial Conditions and Particular Solutions with Description of Behavior: I looked at the two initial conditions: 0 y(0) = ! 2 and y(0) = 1 2 − 14 ! . In the first case, the bob is initially at equilibrium and is pushed upward with a velocity of 2 m/s. The particular solution is y=e ! sin 2t −t/2 2 cos 2t − 1 2 . sin 2t The trajectory is shown in Fig. 2 on the left. The bob initially moves upward above equilibrium. It hits its highest point (of about 0.65 m above equilibrium) when its velocity hits 0. It then moves back toward equilibrium, reaching its largest downward velocity (of about -1.05 m/s) before it hits equilibrium. It then moves past equilibrium, hitting its lowest point (of about 0.35 m) when its velocity hits 0. It moves back toward equilibrium, again hitting a maximum of velocity of about 0.45 m/s) before it reaches equilibrium. It crosses equilibrium and moves up to a maximum of displacment that is much smaller than its previous maximum. It continues oscillating up and down through equilibrium, with its maximum and minimum displacements and velocities in each cycle approacing 0. In the second, case, the bob is initially − 14 1 2 m above equilibrium and is pushed with a downward velocity of m/s. The particular solution is y = e−t/2 1 2 1 4 cos 2t cos 2t − sin 2t ! . The trajectory is shown in Fig. 2 on the right. This trajectory is very similar to the one described previously. The bob starts above equilibrium moving in a downward direction. Just like for the other initial condition, 4 Figure 2. Trajectories for {m = 1 kg, c = 1 N ·s/m, k = 17 4 N/m} Left: Initial displacement is 0 and initial velocity is 2 m/s. Right: Initial displacement is 0.5 m and initial velocity is -0.25 m/s. Figure 3. Phase Portrait for Damped System, {m = 4 kg, c = 12 N · s/m, k = 9 N/m} the bob oscillates up and down through equilibrium and its amplitude of oscillation decreases over time, until it is nearly at rest at the equilibrium point. 3.4. Repeated Eigenvalues. With the parameters {m = 4 kg, c = 12 N · s/m, k = 9 N/m}, the system can be written: 0 y = 0 1 −9/4 −3 ! y. These parameters are realistic, representing a very large strong spring, with a sizeable bob on it. The damping is very strong. The system is critically damped. The general solution to this system can be written: −3t/2 y(t) = e c1 + c2 (1 + t) − 23 c1 + c2 (− 12 − 23 t) 3.4.1. Phase Portrait. Fig. 3 shows a phase portrait for this system. ! . 5 Figure 4. Trajectories for {m = 4 kg, c = 12 N ·s/m, k = 9 N/m} Left: Initial displacement is 2 m below equilibrium and initial velocity is 7 m/s. Right: Initial displacement is 10 m below equilibrium and initial velocity is 10 m/s. 3.4.2. Initial Conditions and Particular Solutions with Description of Behavior: I looked at the two initial conditions: −2 y(0) = ! and 7 −10 y(0) = 10 ! . In the first case, the bob starts 2 m below equilibrium with an upward velocity of 7 m/s. The particular solution is: −3t/2 y=e 4t − 2 ! 7 − 6t . The trajectory is shown in Fig. 4 on the left. It shows that the bob starts below the equilibrium with upward velocity. It moves upward through the equilibrium to its maximum displacement of about 0.46 m when its velocity is 0. Then it moves back down toward equilibrium, with negative velocity, but its velocity approaches 0 and it nevers reaches equilibrium. In the second case, the bob starts 10 m below equilibrium with an upward velocity of 10 m/s. The particular solution is: −5t − 10 y = e−3t/2 10 + ! . 15 2 t The trajectory is shown in Fig. 4 on the right. The bob starts far (10 m) below equilibrium with an upward velocity of 10 m/s. It approaches equilibrium from below as its velocity drops to 0, never reaching equilibrium. 3.5. Non-homogeneous System. 3.5.1. Meaning of Inhomogeneity: I used the same parameters as in my first homogeneous system: {m = 1 kg, c = 1 N · s/m, k = 17/4 N/m}. I chose a sinusoidal external forcing for the inhomogeneity, fext (t) = 16 cos 4t, modeling a motor attached to the bob pushing it up and down smoothly over time. ! ! 0 0 f (t) = = fext (t) 16 cos 4t m The system in matrix-vector form is then: 0 y = 0 1 − 17 4 −1 ! y+ 0 16 cos 4t ! . 6 Figure 5. Trajectories for {m = 1 kg, c = 1 N · s/m, k = 17 4 N/m, fext (t) = 16 cos 4t N } Left: Initial displacement is 2 m below equilibrium and initial velocity is -7 m/s. Right: Initial displacement is 0 and initial velocity is 0. From wolframalpha.com, a particular solution is: yp = 64(16 sin 4t−47 cos 4t) 2465 4t+16 cos 4t) − 256(47 sin2465 ! . Adding this to my general solution to the homogeneous system, the general solution to the non-homogeneous system is: y(t) = 64(16 sin 4t−47 cos 4t) 2465 256(47 sin 4t+16 cos 4t) c2 −t/2 )e sin 2t + 2 2465 c1 e−t/2 cos 2t + c2 e−t/2 sin 2t + (− c21 + 2c2 )e−t/2 cos 2t − (2c1 + ! Fig. 5 shows, on the left, a trajectory satisfying the initial condition ! −2 . y(0) = −7 In this case, the bob starts below equilibrium with a downward velocity. It moves down almost 1 m to its lowest point (when the velocity is 0 as was always the case without forcing). It starts moving back toward equilibrium with positive velocity but before it reaches it, its velocity becomes negative and it starts moving down away from equilibrium again. But then the velocity increases and it begins moving toward equilibrium again. From then on, it moves up and down past equilibrium approaching a perfectly sinusoidal motion (like an undamped system) with angular frequency matching that of the motor, 4 rad/s. The effect of the initial conditions decay, and the system follows the motor. Fig. 5 shows, on the right, a trajectory satisfying the initial condition ! 0 y(0) = . 0 In this case, the bob is released from rest at the equilibrium position. So, the motor has no initial conditions to ‘overcome.’ Thus, the motion just follows the motor from the start; the bob moves in perfect sinusoidal motion with the same frequency as the motor. 4. Appendix Handwritten calculations. Full detail is not required, but it needs to be enough for us to see that you have solved the systems. The calculations can be in done with parameters, rather than specific values.
Surname1 THE RC CONSTANT OF AN ELECTRODE Introduction First-order differential equations help to describe many chemical reactions and other general scientific occurrences. These equations apply to explain the laws of nature. The RC ‘constant’ at a disk electrode is one of the many chemical reactions described by first-order differential equations (Barnartt et al 2010). Electrodes have advanced with time from large sheets of platinum to small inset disks made of golden materials. This evolution has, however, unmatched electrochemical experiments with developments in experimentation, for example, the RC time constant (Haková et al 2012). There is a common assumption that the typical time which dictates the double layers of a disk charged, and consistency accessible in a similar manner, electrodes is a constant. There are many considerations of period nature such as dynamic processes involved in evaluating an investigation of electrochemistry. One of the factors is the timescale regarding the charging of the surface of the electrode through the over-resistance of the cell. The RC constant is the span of the process of charging (Guziejewski 2014). The RC constant works on the assumption that the method of charging is commonly formed on the ground of a capacitor and a resistor in a series pattern. The effect of the two electrical elements creates a product of time. RC time constant also plays a significant role in modeling devices of electrochemical nature (Dunford et al 2013. The charging Process Consider a case of potassium nitrate (KNO3) solution containing only the following molecules and ions: K+, NO3-, H2O, H3O+, OH−. If the metal and solution potential difference at both electrodes is inadequate for an electrochemical reaction, any voltage change between the Surname1 electrodes will result to charging or discharging the double layer of the electrodes (Çelik et al 2012). Therefore, the current will flow through a resistor formed by the electrolyte solution into or from the capacitor resulting from the double layers. Suppose in the beginning the double layer is discharged, that is, the electrode resting at potential zero. In an experiment of a potential step, the double layer is charged, and the capacitor is filled (Murovtsev et al 2014). Initially (t = 0), the current flows through the resistor. After charging the current returns to zero, when the capacitor charges to its potential difference ΔΕ. Fig. 1: The capacitor formed by the double layer of the electrode (C) and the resistor resulting from the electrolyte solution (R) in a cell of the electrochemical reaction. Considering Kirchhoff’s law, the resultant potential across the resistor and the capacitor is the sum of the decline of two potentials (Oldham et al 2013). 𝛥𝛦 = 𝛥𝛦(𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟) + 𝛥𝛦(𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟) (equation 1) The potential decrease across the resistor coincides with Ohm’s law (Petzold 2011): 𝛥𝛦(𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟) = 𝑅𝐼 (equation 2) Where: R = the solution’s resistance; and I = the current The potential decrease across the capacitor is: 𝛥𝛦𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟 = 𝑄 𝐶 (equation 3) Where: Q = the charge; and C = the capacitor’s capacitance, that is, the double layer Replacing the terms from equation 2 and equation 3 as below: ΔΕ – RI To find I; 𝑄 𝐶 =0 (equation 4) Surname1 I= ΔΕ Q - 𝑅 (equation 5) 𝑅𝐶 The current being the first derivative of charge against time (𝐼 = 𝛥𝑄/𝛥𝑡), equation 5 can be rewritten as follows: Δq ΔΕ q 1 (q – C*ΔΕ) = Δq + 𝑅𝐶 (q – C*ΔΕ) = 0 Δt - =– Δt 𝑅 𝑅𝐶 𝑅𝐶 (equation 6) 1 (equation 7) Equation 7 is a nonhomogeneous linear first order equation If equation 7 is integrated in the limits of 0 and q, and 0 and t, the result is: Δq 𝑞 ∫0 (q – C∗ΔΕ) =− 1 𝑅𝐶 𝑞 ∫0 Δt = − t 𝑅𝐶 (equation 8) At t = 0 and q = 0 from the basis that the electrode is at zero potential (zero charge) To solve the left side integral, we substitute; Q’ = q – C*ΔΕ (equation 9) Therefore, q = q’ + C*ΔΕ and 𝛥𝑞 = 𝛥𝑞’ q’t Δq′ ∫q’t = 0 q =− t (equation 10) 𝑅𝐶 NOTE: This can be equated to solving the differential equation: 1nq'qtqt=0 =− Δq′ Δt + 1 RC 𝑞′ = 0 (equation 11) t RC By substituting back q’ = q – C*ΔΕ and Δq′ = Δq and then describing 𝑞’𝑡 = 0 = – C*ΔΕ which equates to the condition 𝑞’𝑡 = 0 = 0 resulting: t 1n (q – C*ΔΕ) - 1n (– C*ΔΕ) = - RC (equation 12) 𝑞−𝐶 ∗ ΔΕ 1n ( −𝐶 ∗ ΔΕ ) = − 𝑞−𝐶 ∗ ΔΕ t RC – ( −𝐶 ∗ ΔΕ ) = e-t/RC (equation 13) (equation 14) Surname1 q – C*ΔΕ = – C*ΔΕ * e-t/RC (equation 15) q = – C*ΔΕ * e-t/RC + C*ΔΕ (equation 16) q = – C*ΔΕ (1 – e-t/RC) (equation 17) Referring 𝐼 = 𝛥𝑄/𝛥𝑡, it then derives from equation 17 that: I= Δq Δt =– 1 𝑅𝐶 (– C*ΔΕ) * e-t/RC = Remember, the ΔE 𝑅 ΔE 𝑅 -t/RC *e (equation 18) should be equal to the current at the start I0 at t = 0, equation 18 results to a current transient: I = I0 * e-t/RC (equation 19) which explains the exponential decay of the current during the potential step. RC is also referred to as the time constant Ƭ and it is expressed as: I = I0 * e-t/Ƭ (equation 20) Alternatively, nonhomogeneous first-order differential equation 7 can be solved by first solving the homogeneous equation Δqh Δt 1 + 𝑅𝐶 qh(t) = 0 (equation 21) Resulting to 𝑞ℎ (𝑡) as expressed in an exponential phase. Likewise, a solution 𝑞𝑝 (𝑡), which is a nonhomogeneous differential equation is solved by substituting: 𝑞𝑝 (𝑡) = 𝑢 (𝑡) ∗ 𝑞ℎ (𝑡) (equation 22) and its first derivative: 𝑞’𝑝 (𝑡) = 𝑢’(𝑡) ∗ 𝑞ℎ (𝑡) + 𝑢 (𝑡) 𝑞’ℎ (𝑡) (equation 23) The two expressions for 𝑞𝑝 (𝑡) and its first derivative 𝑞’𝑝 (𝑡) are substituted by equation 7, resulting to finding u (t) and 𝑞𝑝 (𝑡). Therefore, the full answer to equation 7 is the answer of 𝑞ℎ (𝑡) (homogeneous differential equation) added to the solution 𝑞𝑝 (𝑡): 𝑞 (𝑡) = 𝑞ℎ (𝑡) + 𝑞𝑝 (𝑡) (equation 24) Surname1 In conclusion, in most investigations it is important to ascertain the RC time constant value of the electrode/electrolyte set-up, such that the charging current is void of faradaic currents under experimentation (Tsirulik 2013) Surname1 Works Cited Barnartt, Sidney, and Charles A. Johnson. "Electrode Kinetics at Constant Potential." Transactions of The Faraday Society, vol. 63, 2010, p. 431. Royal Society of Chemistry (RSC), doi:10.1039/tf9676300431. Dunford, Sean. "Calculating The Time Constant of an RC Circuit." Undergraduate Journal of Mathematical Modeling: One + Two, vol. 2, no. 2, 2013. University of South Florida Libraries, doi:10.5038/2326-3652.2.2.3. Guziejewski, Dariusz et al. "Measuring The Electrode Kinetics of Surface-Confined Electrode Reactions at A Constant Scan Rate." Electroanalysis, vol. 27, no. 1, 2014, pp. 67-73. Wiley, doi:10.1002/elan.201400349. Haková, Alžběta, and Olga Krupková. "Varying First-Order Partial Differential Equations." Journal of Differential Equations, vol. 191, no. 1, 2012, pp. 67-89. Elsevier BV, doi:10.1016/s0022-0396(02)00160-2. Çelik, Ercan et al. "Numerical Method to Solve Chemical Differential-Algebraic Equations."International Journal of Quantum Chemistry, vol. 89, no. 5, 2012, pp. 447-451. Wiley, doi:10.1002/qua.10305. Murovtsev, A. N. "Two-Sided Solutions of Linear Non-Autonomous Homogeneous FunctionalDifferential Equations." Differential Equations, vol. 41, no. 10, 2014, pp. 1417-1424. Springer Nature, doi:10.1007/s10625-005-0293-0. Oldham, K. "The RC Time “Constant” At A Disk Electrode." Electrochemistry Communications, vol. 6, no. 2, 2013, pp. 210-214. Elsevier BV, doi:10.1016/j.elecom.2003.12.002. Surname1 Summers, D., and J.M.W. Scott. "Systems Of First-Order Chemical Reactions." Mathematical And Computer Modelling, vol 10, no. 12, 1988, pp. 901-909. Elsevier BV, doi:10.1016/08957177(88)90182-3. Petzold, Linda et al. "Sensitivity Analysis of Differential-Algebraic Equations and Partial Differential Equations." Computers & Chemical Engineering, vol. 30, no. 10-12, 2011, pp. 1553-1559. Elsevier BV, doi:10.1016/j.compchemeng.2006.05.015. Tsirulik, V. G. "On The Method of Indeterminate Coefficients for Nonhomogeneous Linear Ordinary Differential Equations." Differential Equations, vol. 39, no. 11, 2013, pp. 16571660. Springer Nature, doi:10.1023/b:dieq.0000019362.39546.aa.

Tutor Answer

(Top Tutor) Studypool Tutor
School: UIUC
Studypool has helped 1,244,100 students
flag Report DMCA
Similar Questions
Hot Questions
Related Tags

Brown University





1271 Tutors

California Institute of Technology




2131 Tutors

Carnegie Mellon University




982 Tutors

Columbia University





1256 Tutors

Dartmouth University





2113 Tutors

Emory University





2279 Tutors

Harvard University





599 Tutors

Massachusetts Institute of Technology



2319 Tutors

New York University





1645 Tutors

Notre Dam University





1911 Tutors

Oklahoma University





2122 Tutors

Pennsylvania State University





932 Tutors

Princeton University





1211 Tutors

Stanford University





983 Tutors

University of California





1282 Tutors

Oxford University





123 Tutors

Yale University





2325 Tutors