Section 4.9; Section 5.6. June 30, Free Mechanical Vibrations/Couple Mass-Spring System

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1 Section 4.9; Section 5.6 Free Mechanical Vibrations/Couple Mass-Spring System June 30, 2009

2 Today s Session

3 Today s Session A Summary of This Session:

4 Today s Session A Summary of This Session: (1) Free Mechanical Vibration (no forcing term).

5 Today s Session A Summary of This Session: (1) Free Mechanical Vibration (no forcing term). (2) Coupled Mass-Spring systems

6 Today s Session A Summary of This Session: (1) Free Mechanical Vibration (no forcing term). (2) Coupled Mass-Spring systems (3) Our first exposure to systems of differential equations

7 Today s Session A Summary of This Session: (1) Free Mechanical Vibration (no forcing term). (2) Coupled Mass-Spring systems (3) Our first exposure to systems of differential equations (4) Eigenvalues and Eigenvectors

8 Today s Session A Summary of This Session: (1) Free Mechanical Vibration (no forcing term). (2) Coupled Mass-Spring systems (3) Our first exposure to systems of differential equations (4) Eigenvalues and Eigenvectors

9 Free Mechanical Vibrations Free mechanical vibration=no forcing function, so f (t) = 0. We are dealing with my + by + ky = 0 where m = mass attached to a spring of stiffness k, subject to friction (or damping) proportional to speed with damping constant b.

10 Free Mechanical Vibrations Free mechanical vibration=no forcing function, so f (t) = 0. We are dealing with my + by + ky = 0 where m = mass attached to a spring of stiffness k, subject to friction (or damping) proportional to speed with damping constant b. Four cases: (1) undamped free case; (2) underdamped case; (3) overdamped case; (4) critical damped case.

11 Free Mechanical Vibrations Free mechanical vibration=no forcing function, so f (t) = 0. We are dealing with my + by + ky = 0 where m = mass attached to a spring of stiffness k, subject to friction (or damping) proportional to speed with damping constant b. Four cases: (1) undamped free case; (2) underdamped case; (3) overdamped case; (4) critical damped case.

12 Case 1: Undamped mass-spring system: b = 0. The equation is given by: my + ky = 0

13 Case 1: Undamped mass-spring system: b = 0. The equation is given by: my + ky = 0 or y + k m y = 0

14 Case 1: Undamped mass-spring system: b = 0. The equation is given by: or my + ky = 0 y + k m y = 0 Let ω 2 = k m. The quantity ω = k m is called the angular frequency (measured in radians per second)

15 Case 1: Undamped mass-spring system: b = 0. The equation is given by: or my + ky = 0 y + k m y = 0 Let ω 2 = k m. The quantity ω = k m is called the angular frequency (measured in radians per second) Period: T = 2π ω (measured in seconds)

16 Case 1: Undamped mass-spring system: b = 0. The equation is given by: or my + ky = 0 y + k m y = 0 Let ω 2 = k m. The quantity ω = k m is called the angular frequency (measured in radians per second) Period: T = 2π ω (measured in seconds) Frequency: f = 1 T = ω 2π (measured in Hertz=1/seconds=# cycles per second)

17 Case 1: Undamped mass-spring system: b = 0. The equation is given by: or my + ky = 0 y + k m y = 0 Let ω 2 = k m. The quantity ω = k m is called the angular frequency (measured in radians per second) Period: T = 2π ω (measured in seconds) Frequency: f = 1 T = ω 2π (measured in Hertz=1/seconds=# cycles per second) The solution is given by: y = C 1 cos ωt +C 2 sinωt = A sin(ωt +φ).

18 Undamped case, cont d Here A is called the amplitude, and φ is called the phase. They are given by A = C1 2 + C2 2 φ = arctan C 1. C 2

19 Undamped case, cont d Here A is called the amplitude, and φ is called the phase. They are given by A = C1 2 + C2 2 φ = arctan C 1. C 2 Example A 1/8 kg manss is attached to a spring with stiffness k = 16N/m. The mass is displaed 0.5 m to the right of the equilibrium point and given an outward initial velocity of 2 m/s. (a) Neglecting damping, find a formula of the displacement as a function of time. Display the values of ω, T, f, A and φ.

20 Undamped case, cont d Here A is called the amplitude, and φ is called the phase. They are given by A = C1 2 + C2 2 φ = arctan C 1. C 2 Example A 1/8 kg manss is attached to a spring with stiffness k = 16N/m. The mass is displaed 0.5 m to the right of the equilibrium point and given an outward initial velocity of 2 m/s. (a) Neglecting damping, find a formula of the displacement as a function of time. Display the values of ω, T, f, A and φ. (b) How long after release does the mass pass first through the equilibrium position?

21 Undamped case, cont d Answers: (a) ω = 8 2 rad/sec; f = 1.8 cycles/sec (or Hz); T = sec.

22 Undamped case, cont d Answers: (a) ω = 8 2 rad/sec; f = 1.8 cycles/sec (or Hz); T = sec. y(t) = 1 2 cos 8 2t sin8 2t

23 Undamped case, cont d Answers: (a) ω = 8 2 rad/sec; f = 1.8 cycles/sec (or Hz); T = sec. A = y(t) = 1 2 cos 8 2t sin8 2t ( 1 2 )2 + ( 1 8 )2 = 17 8 and tanφ = 1/2 1/8 = 4 so φ = rad. y(t) = 17 8 sin(8 2t )

24 Undamped case, cont d Answers: (a) ω = 8 2 rad/sec; f = 1.8 cycles/sec (or Hz); T = sec. A = y(t) = 1 2 cos 8 2t sin8 2t ( 1 2 )2 + ( 1 8 )2 = 17 8 and tanφ = 1/2 1/8 = 4 so φ = rad. y(t) = 17 8 sin(8 2t ) (b) y(t) = 0 means sin(8 2t + φ) = 0 so 8 2t + φ = kπ for k integer. Solving gives: t = kπ φ ω = 0.16 sec. Every 1/2 period (or 0.28 sec) the mass goes through the equilibrium point.

25 Overdamped case: b 2 4k m > 0 The equation is my + by + ky = 0

26 Overdamped case: b 2 4k m > 0 The equation is my + by + ky = 0 The characteristic equation is given by mr 2 + br + k = 0

27 Overdamped case: b 2 4k m > 0 The equation is my + by + ky = 0 The characteristic equation is given by mr 2 + br + k = 0 so r = b b 2m ± 2 4k m 2m The solution is given by: y(t) = C 1 e r1t + C 2 e r 2t

28 Overdamped case: b 2 4k m > 0 The equation is my + by + ky = 0 The characteristic equation is given by mr 2 + br + k = 0 so r = b b 2m ± 2 4k m 2m The solution is given by: y(t) = C 1 e r 1t + C 2 e r 2t Note that both r 1 and r 2 are negative (why?), so as t, y(t) 0.

29 Underdamped case: b 2 4k m < 0 With ω = 4k m b 2 2m, the roots of the characteristic equation are: r = b 2m ± i ω So the solution is given by: y(t) = C 1 e b 2m t cos ωt + C 2 e b 2m t sinωt.

30 Underdamped case: b 2 4k m < 0 With ω = 4k m b 2 2m, the roots of the characteristic equation are: r = b 2m ± i ω So the solution is given by: y(t) = C 1 e b 2m t cos ωt + C 2 e b 2m t sinωt. Over time y(t) dies in an oscillatory fashion. ω is a quasi-angular frequency, T = 2π/ω is called the quasiperiod and f = 1/T is called the quasi-frequency.

31 Underdamped case: b 2 4k m < 0 With ω = 4k m b 2 2m, the roots of the characteristic equation are: r = b 2m ± i ω So the solution is given by: y(t) = C 1 e b 2m t cos ωt + C 2 e b 2m t sinωt. Over time y(t) dies in an oscillatory fashion. ω is a quasi-angular frequency, T = 2π/ω is called the quasiperiod and f = 1/T is called the quasi-frequency. The solution can be put in the form where (as before): A = y(t) = A e b 2m t sin(ωt + φ) C C2 2 φ = arctan C 1 C 2.

32 Underdamped case: b 2 4k m < 0 With ω = 4k m b 2 2m, the roots of the characteristic equation are: r = b 2m ± i ω So the solution is given by: y(t) = C 1 e b 2m t cos ωt + C 2 e b 2m t sinωt. Over time y(t) dies in an oscillatory fashion. ω is a quasi-angular frequency, T = 2π/ω is called the quasiperiod and f = 1/T is called the quasi-frequency. The solution can be put in the form where (as before): A = y(t) = A e b 2m t sin(ωt + φ) C C2 2 φ = arctan C 1 C 2.

33 Critically damped case: b 2 4k m = 0 In this case, the characteristic equation has a double root r = b 2m, So the solution is y(t) = C 1 e b 2m t + C 2 t e b 2m t

34 Critically damped case: b 2 4k m = 0 In this case, the characteristic equation has a double root r = b 2m, So the solution is y(t) = C 1 e b 2m t + C 2 t e b 2m t Example: (related to webassign question and example p. 234)

35 Critically damped case: b 2 4k m = 0 In this case, the characteristic equation has a double root r = b 2m, So the solution is y(t) = C 1 e b 2m t + C 2 t e b 2m t Example: (related to webassign question and example p. 234) Find the value of b for which y + b y + 25y = 0 y(0) = 1, y (0) = 0 is critically damped. Solve for y(t) in this case and sketch it.

36 Critically damped case: b 2 4k m = 0 In this case, the characteristic equation has a double root r = b 2m, So the solution is y(t) = C 1 e b 2m t + C 2 t e b 2m t Example: (related to webassign question and example p. 234) Find the value of b for which y + b y + 25y = 0 y(0) = 1, y (0) = 0 is critically damped. Solve for y(t) in this case and sketch it. Answer: b = 10; y(t) = (1 t)e 5t

37 5.6: Coupled Mass-Spring System: Example 1, p. 308 See the graph and description of the couple mass-spring system on p. 308.

38 5.6: Coupled Mass-Spring System: Example 1, p. 308 See the graph and description of the couple mass-spring system on p I will show the derivation of the following equations in class:

39 5.6: Coupled Mass-Spring System: Example 1, p. 308 See the graph and description of the couple mass-spring system on p I will show the derivation of the following equations in class: We wish to solve: { m1 x = k 1 x + k 2 (y x) m 2 y = k 2 (y x)

40 5.6: Coupled Mass-Spring System: Example 1, p. 308 See the graph and description of the couple mass-spring system on p I will show the derivation of the following equations in class: We wish to solve: { m1 x = k 1 x + k 2 (y x) m 2 y = k 2 (y x) In our example: m 1 = 2kg ; k 1 = 4N/m; m 2 = 1kg ; and k 2 = 2N/m. So: { 2x = 6x + 2y y = 2x 2y

41 5.6: Coupled Mass-Spring System: Example 1, p. 308 See the graph and description of the couple mass-spring system on p I will show the derivation of the following equations in class: We wish to solve: { m1 x = k 1 x + k 2 (y x) m 2 y = k 2 (y x) In our example: m 1 = 2kg ; k 1 = 4N/m; m 2 = 1kg ; and k 2 = 2N/m. So: { 2x = 6x + 2y y = 2x 2y

42 Example 1, p. 308, cont d Let s clean this up a little bit: { x = 3x + y y = 2x 2y

43 Example 1, p. 308, cont d Let s clean this up a little bit: { x = 3x + y y = 2x 2y This is a linear system of differential equations of second order (which is homogeneous; why?) We can put it in the form: { (D 2 + 3)x y = 0 2x + (D 2 + 2)y = 0

44 Example 1, p. 308, cont d Let s clean this up a little bit: { x = 3x + y y = 2x 2y This is a linear system of differential equations of second order (which is homogeneous; why?) We can put it in the form: { (D 2 + 3)x y = 0 2x + (D 2 + 2)y = 0 Therefore (D 2 + 2)(D 2 + 3)x (D 2 + 2)y = 0

45 Example 1, p. 308, cont d Let s clean this up a little bit: { x = 3x + y y = 2x 2y This is a linear system of differential equations of second order (which is homogeneous; why?) We can put it in the form: { (D 2 + 3)x y = 0 2x + (D 2 + 2)y = 0 Therefore (D 2 + 2)(D 2 + 3)x (D 2 + 2)y = 0 We note that: (D 2 + 2)y = 2x.

46 Example 1, p. 308, cont d Let s clean this up a little bit: { x = 3x + y y = 2x 2y This is a linear system of differential equations of second order (which is homogeneous; why?) We can put it in the form: { (D 2 + 3)x y = 0 2x + (D 2 + 2)y = 0 Therefore (D 2 + 2)(D 2 + 3)x (D 2 + 2)y = 0 We note that: (D 2 + 2)y = 2x. So: (D 2 + 2)(D 2 + 3)x 2x = 0.

47 Example 1, p. 308, cont d Let s clean this up a little bit: { x = 3x + y y = 2x 2y This is a linear system of differential equations of second order (which is homogeneous; why?) We can put it in the form: { (D 2 + 3)x y = 0 2x + (D 2 + 2)y = 0 Therefore (D 2 + 2)(D 2 + 3)x (D 2 + 2)y = 0 We note that: (D 2 + 2)y = 2x. So: (D 2 + 2)(D 2 + 3)x 2x = 0. We clean this up a little bit: (D 4 + 5D 2 + 4)x = 0.

48 Example 1, p. 308, cont d Let s clean this up a little bit: { x = 3x + y y = 2x 2y This is a linear system of differential equations of second order (which is homogeneous; why?) We can put it in the form: { (D 2 + 3)x y = 0 2x + (D 2 + 2)y = 0 Therefore (D 2 + 2)(D 2 + 3)x (D 2 + 2)y = 0 We note that: (D 2 + 2)y = 2x. So: (D 2 + 2)(D 2 + 3)x 2x = 0. We clean this up a little bit: (D 4 + 5D 2 + 4)x = 0.

49 Example 1, p. 308, cont d Then factor: (D 2 + 1)(D 2 + 4)x = 0

50 Example 1, p. 308, cont d Then factor: (D 2 + 1)(D 2 + 4)x = 0 Therefore: x(t) = C 1 cos t + C 2 sint + C 3 cos 2t + C 4 sin2t and y = (D 2 + 3)x = 2C 1 cos t + 2C 2 sint C 3 cos 2t C 4 sin2t

51 Example 2 Use the method of the previous example (the annihilator method) to solve the first order (homogeneous) linear system: { x = 3x + 4y y = 4x + 3y

52 Example 2 Use the method of the previous example (the annihilator method) to solve the first order (homogeneous) linear system: { x = 3x + 4y y = 4x + 3y Answer: x(t) = C 1 e 7t + C 2 e t and y(t) = C 1 e t C 2 e t

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