Introductory Physics. Week 2015/05/29

Transcription

1 2015/05/29

2 Part I Summary of week 6

3 Summary of week 6 We studied the motion of a projectile under uniform gravity, and constrained rectilinear motion, introducing the concept of constraint force. Then we introduced the concept of work, kinetic energy, and potential energy. We used them to analyse the vertical motion under uniform gravity.

4 Projectile without air resistance Question 3 A particle of mass m is projected with speed v 0 in a direction making an angle θ with the horizontal. The particle is under uniform gravity, and no drag force. Find the subsequent motion. Take Cartesian coordinate system (x, z). The equation of motion is m d v dt = mg e z, with the initial condition v = v 0 cos θ e x + v 0 sin θ e z at t = 0.

5 Projectile without air resistance In Cartesian coordinate system, we can analyse the x and z component of vector independently. This is because e x = e z = 0. r = x e x + z e z r = ẋ e x + ż e z r = ẍ e x + z e z

6 Projectile without air resistance The equation of motion m dv dt = mg e z We obtain the two scalar equation of motion for each x and z component. m dv x dt = 0, Using the initial conditions, v x = v 0x = v 0 cos θ, x = (v 0 cos θ)t, mdv z dt = mg v z = v 0z gt = v 0 sin θ gt z = (v 0 sin θ)t 1 2 gt2

7 the path of the projectile To find the path of the projectile, we eliminate t from the equations x = (v 0 cos θ)t, t = x v 0 cos θ z = (tan θ)x g 2v 2 0 cos 2 θ x2 This is an inverted parabola. z = (v 0 sin θ)t 1 2 gt2

8 the Monkey and the Hunter

9 the Monkey and the Hunter

10 the Monkey and the Hunter

11 the Monkey and the Hunter Let us write r = x e x + y e y Then because e x = e y = 0, (just as in the case of Cartesian coordinate system) r = ẋ e x + ẏ e y, r = ẍ ex + ÿ e y This means we can independently consider the change of x and y.

12 the Monkey and the Hunter The equations of motion for the bullet are ẍ P = 0, ÿ P = g with initial conditions of ẋ P = v 0, x P = 0, ẏ P = 0, y P = 0 The equations of motion for the monkey are ẍ M = 0, ÿ M = g with initial conditions of ẋ M = 0, x M = l, ẏ M = 0, y M = 0

13 the Monkey and the Hunter Because both the differential equations for y P and y M are same (ÿ P = ÿ M = g), and their inital conditions are same (ẏ P = ẏ M = 0, y P = y M = 0), It is obvious that y P = y M holds for any t. Since x P = v 0t and x M = l, the bullet hit the monkey (x P becomes equal to x M ) when t = l/v 0.

14 Constraint force Constraint forces are forces which are not prescribed beforehand, but are sufficient to enforce a specified geometrical constraint. The examples A particle sliding on a smooth fixed slope. The normal reaction force exarted on the particle by the slope is a constraint force. A particle is connected to a fixed point by an inextensible string. When the string is taut, the distance between the particle and the fixed point is kept constant. The tention exarted on the particle by the string is a constraint force.

15 Work, kinetic energy Definition work done by force F W = kinetic energy t2 t 1 F vdt K = 1 m v v 2 Energy Principle K 2 K 1 = W The increase of the kinetic energy in a given time interval is equal to the total work done by applied force(s) in the time interval.

16 Potential energy and energy conservation in rectilinear motion We consider a case where a particle P moves along x-axis under the force field F (x). If F (x) can be written as F (x) = potential energy. dv (x), V (x) is the dx Energy conservation in rectilinear motion When a particle undergoes rectilinear motion in a force field, the mechanical energy (the sum of its kinetic and potential energy) remains constant in the motion.

17 Part II Energy and Work (Cont.)

18 Work done by constraint force In most of idealized problems, the constraint force is perpendicular to the velocity of the particle ( F v = 0). In this case, the work done by the constraint force is zero. Energy conservation in constrained motion When a particle moves under conservative force, nd is subject to constraint forces that do not work, the mechanical energy (the sum of its kinetic and potential energy) remains constant in the motion.

19 Example: block on a slope The surface is smooth, and the block is at rest when t = 0. When the block slides down by l, what is its velocity? Since the normal reaction force does no work, and there is no friction force, the energy conservation applies in the system. Let us take the reference point to the original position = 1 2 mv2 mgh = 1 2 mv2 mgl sin θ v = 2gl sin θ

20 Example: block on a slope The surface is smooth, and the block was at rest when t = 0. When the block slides down by l, what is its velocity? The equation of motion is m d v dt = mg e z + N e z Noting e z = sin θ e x + cos θ e z, we obtain m dv dt = mg sin θ 0 = N mg cos θ

21 Example: block on a slope v = (g sin θ)t x = 1 (g sin θ)t2 2 when t = t l, the block arrived at x = l l = 1 2 (g sin 2l θ)t2 l t l = g sin θ 2l v l = (g sin θ) g sin θ = 2gl sin θ where h = l sin θ is the change of the height of the block.

22 Example: Downhill Skier The constraint force is often not known beforehand, so we often face the case when we can not solve the equation of motion. But we can extract some useful information using the energy principle (and energy conservation). When a skier descends on a bumpy slope, the magnitude and the direction of the normal reaction force changes so that the skier is kept on the slope. But when the slope is smooth, the constraint force does no work, so the energy conservation applies. The speed of the skier after losing attitude h is

23 Part III Linear oscillations

24 Simple harmonic motion Let s consider the motion a particle attached to a spring, whose restoring force F (x) is given by Hook s law (F (x) = kx). We assume there are no registance nor other external forces. x kx

25 Simple harmonic motion The equation of motion is mẍ + kx = 0 ẍ + k m x = 0 How to solve this differential equation? where k and m are positive constants.

26 differential equation formulas formula The general solution of differential equation is y = C 1 e λ 1t + C 2 e λ 2t ÿ + a 1 ẏ + a 2 y = 0 (a 2 1 4a 2 0) y = (C 1 + C 2 t)e λt (a 2 1 4a 2 = 0),where λ is solution(s) of a quadratic equation λ 2 + a 1 λ + a 2 = 0 and C 1 and C 2 are the integral constants.

27 differntial equation formulas formula When a 2 1 4a 2 < 0, λ = µ ± iω (µ and ω are real numbers). Then it is often convenient to write the general solution as y = e µt (A cos ωt + B sin ωt) = Ce µt cos(ωt + δ), where A, B, C, δ are the (real) integration constants. Euler s formula e iθ = cos θ + i sin θ

28 Simple harmonic motion The equation of motion is ẍ + k m x = 0ẍ + ω2 x = 0, where ω = In this case, the general solution is x = A cos ωt + B sin ωt = C cos(ωt + δ) k m This is called simple harmonic motion (SHM). The particle makes infinitely many oscillations of constant amplitude C. The period of oscillation is T = 2π ω. If x = x 0 and v = 0 when t = 0, x = x 0 cos ωt

29 Simple harmonic motion x 0 = 1, ω =

30 Damped simple harmonic motion When there is a linear resistance on the particle, the equation of motion is mẍ = kx bẋ mẍ + bẋ + kx = 0 (b > 0, k > 0) This motion is called damped simple harmonic motion. x bv v kx

31 Damped simple harmonic motion The particle is released from x = x 0 when t = 0. The particle starts moving, and after long period of time, the particle will be at rest at x = 0. We want to make this damping as fast as possible. What should we do? (1) make linear resistance as large as possible (2) make linear resistance to an optimum value (3) make linear resistance as small as possible (4) the magnitude of linear resistance does not affect the time for oscillation to be damped to zero

32 Damped simple harmonic motion The equation of motion is mẍ + bẋ + kx = 0 ẍ + 2γẋ + ω0x 2 = 0 (γ = b k 2m, ω 0 = m ) The behavior of the solution depends whether γ > ω 0, γ = ω 0 or γ < ω 0.

33 case 1: γ < ω 0 In this case, the quadratic equation λ 2 + 2γλ + ω 2 0 = 0 have two complex solutions, λ = γ ± iω, where ω = ω 2 0 γ 2. The solution of the differntial equation is x = e γt (A cos ωt + B sin ωt), where A and B are real constants. This is called under-damped simple harmonic motion.

34 case 2: γ > ω 0 In this case, the quadratic equation λ 2 + 2γλ + ω 2 0 = 0 have two real solutions, λ = γ ± δ, where δ = γ 2 ω 2 0. The solution of the differntial equation is x = Ae ( γ+δ)t + Be ( γ δ)t = e γt (Ae δt + Be δt ), where A and B are real constants. This is called over-damped simple harmonic motion.

35 case 3: γ = ω 0 In this case, the quadratic equation λ 2 + 2γλ + ω 2 0 = 0 have a double root, λ = γ. The solution of the differential equation is x = (A + Bt)e γt, where A and B are real constants. This is called critically-damped simple harmonic motion.

36 Damped simple harmonic motion If x = x 0 and ẋ = 0 at t = 0, x = x 0 e γt (cos ωt + γ ω sin ωt) ω = ω 2 0 γ 2 x = x 0 e γt (1 + γt) γ = ω 0 x = x 0 2 e γt ((1 + γ δ )eδt + (1 γ δ )e δt ) δ = γ 2 ω0 2

37 Damped simple harmonic motion ω 0 = 4, γ = 0 (no damping)

38 Damped simple harmonic motion ω 0 = 4, γ = 0,

39 Damped simple harmonic motion ω 0 = 4, γ = 0, 0.5,

40 Damped simple harmonic motion ω 0 = 4, γ = 0, 0.5,

41 Damped simple harmonic motion ω 0 = 4, γ = 0, 0.5, 1.5,

42 Damped simple harmonic motion ω 0 = 4, γ = 0, 0.5, 1.5, 3.0,

43 Damped simple harmonic motion ω 0 = 4, γ = 0, 0.5, 1.5, 3.0, 4.0, 6.0,

44 Damped simple harmonic motion The particle is released from x = x 0 when t = 0. The particle starts moving, and after long period of time, the particle will be at rest at x = 0. We want to make this damping as fast as possible. What should we do? (1) make linear resistance as large as possible (2) make linear resistance to an optimum value (3) make linear resistance as small as possible (4) the magnitude of linear resistance does not affect the time for oscillation to be damped to zero

45 Damped simple harmonic motion The particle is released from x = x 0 when t = 0. The particle starts moving, and after long period of time, the particle will be at rest at x = 0. We want to make this damping as fast as possible. What should we do? (1) make linear resistance as large as possible (2) make linear resistance to an optimum value (3) make linear resistance as small as possible (4) the magnitude of linear resistance does not affect the time for oscillation to be damped to zero

46 (brief introduction) Detailed mathematical explanation will be skipped in this lecture. I ll give a brief introduction of.

47 (brief introduction) When there is an external driving force acting on the linear oscillator, the behavior of the oscillator is called driven motion. The phenomenon of resonance appears in the driven motion, and you will encounter this phenomenon in many other situations. A swing (hanging seat) on the playing field A tuning circuit in a radio A musical instrument Absorption spectrum of atoms and molecules.

48 Tacoma Narrow Bridge Tacoma Narrow Bridge was completed on 1st July, It collapsed on 7th November, 1940.

49 (brief introduction) When the external force is time harmonic, the equation of motion will be mẍ + bẋ + kx = f 0 cos(pt) So we consider (b > 0, k > 0, f 0 is amplitude of the force) ẍ + 2γẋ + ω 2 0x = F 0 cos(pt)

50 resonance (brief introduction) ẍ + 2γẋ + ω 2 0x = F 0 cos(pt) The solution of this differential equation is x = A cos(pt δ) + x damped where x damped is the general solution of ẍ + 2γẋ + ω 2 0x = 0 (damped oscillation) This means that after a sufficiently long time, x a cos(pt δ).

51 Resonance Simple harmonic motion ẍ + 2γẋ + ω 2 0x = F 0 cos(pt)x = A cos(pt δ) + x damped The driving force with angular frequency p will induce large amplitude A in the system when p is close to ω 0, if damping is not too large. This is called resonance.

52 resonance (brief introduction) We set ω 0 = 1. For γ = 1, 0.5, 0.2, 0.1, 0.02, gamma = 1, omega_0 = 1 gamma = 0.5, omega_0 = 1 gamma = 0.2, omega _ 0 = 1 gamma = 0.1, omega_0 = 1 gamma=0.02, omega_0 = You will have infinetely large amplitude at p = ω 0 if there is no registance (γ = 0).

53 Collapse of Tacoma Narrow Bridge Tacoma Narrow Bridge was completed on 1st July, It collapsed on 7th November, The cause of the collapse is said to be aeroelastic fluttering. You see on the video that the bridge vibrate torsionally - there was little damping mechanism on the bridge to suppress this vibration. When we build a bridge or a building, we need to analyse its vibration frequencies and make sure there are enough damping for each vibration.

54 Coupled oscillations

55 Coupled oscillations (brief introduction) What happens if there are two SHMs are coupled with another spring? The equations of motion are mẍ = kx k (x y) = (k + k )x + k y mÿ = k (y x) ky = k x (k + k )y This is a set of simultaneous differential equations (a system of differntial equations).

56 Coupled oscillations (brief introduction) The set of simultaneous differential equations mẍ = kx k (x y) = (k + k )x + k y mÿ = k (y x) ky = k x (k + k )y can be written as d 2 dt 2 ( ) x = y where a = k + k m, b = k m. ( a b b a ) ( x y )

57 Coupled oscillations (brief introduction) This differential equations can be easily solved if we can find a matrix U such that ( ) ( ) a b λ1 0 = U U 1 b a 0 λ 2 This is called diagonalization of matrix.

58 Coupled oscillations (brief introduction) ( ) ( ) ( ) ( ) ( ) d 2 x a b x λ1 0 x = = U U 1 dt 2 y b a y 0 λ 2 y ( ) ( ) ( ) d 2 x dt U 1 λ1 0 x = U 1 2 y 0 λ 2 y ( ) ( ) ( ) d 2 p λ1 0 p = dt 2 q 0 λ 2 q,where p = λ 1 p q = λ 2 q ( ) ( ) p x = U 1 q y

59 Coupled oscillations (brief introduction) In this example, U = U 1 = 1 ( ) and λ 1 = a b = k m, λ 2 = a + b = k + 2k m. Note: λ are eigenvalues of the matrix. ( ) ( ) a b = (a b) 1 ( ) 1 b a ( ) ( ) a b = (a + b) 1 ( ) 1 b a 1 2 1

60 Coupled oscillations (brief introduction) Using this matrix U, ( ) ( ) ( ) d 2 x a b x = dt 2 y b a y ( ) ( ) ( ) ( ) ( ) d x λ x 2 = 2 dt y 0 λ y ( ) ( ) ( ) d 2 x + y λ1 0 x + y = dt 2 x y 0 λ 2 x y

61 Coupled oscillations (brief introduction) Subsituting p = x + y, q = x y, p = λ 1 p = k m p q = λ 2 q = k + 2k m q p = A cos(ω 1 t + γ 1 ), q = B cos(ω 2 t + γ 2 ) k k + 2k where ω 1 = m, ω 2 = m. These are called normal modes of the oscillation system.

62 Coupled oscillations (brief introduction) The motion of particles are superposition of normal modes. x = 1 2 (A cos(ω 1t + γ 1 ) + B cos(ω 2 t + γ 2 )) y = 1 2 (A cos(ω 1t + γ 1 ) B cos(ω 2 t + γ 2 )) k k + 2k where ω 1 = m, ω 2 = m.

63 Coupled oscillations (brief introduction) The motion of particles are superposition of normal modes. x = 1 2 (A cos(ω 1t + γ 1 ) + B cos(ω 2 t + γ 2 )) y = 1 2 (A cos(ω 1t + γ 1 ) B cos(ω 2 t + γ 2 )) k k + 2k where ω 1 = m, ω 2 = m. When difference between ω 1 and ω 2 are small (in this case when k is small), we observe beat on the amplitude of the oscillation.

64 Beat Simple harmonic motion

65 Beat Simple harmonic motion

66 Beat Simple harmonic motion

67 Beat Simple harmonic motion

68 Coupled oscillations (brief introduction) There are many examples of coupled oscillators. molecules crystals bridges and buildings Linear algebra (matrix, eigenvalue, eigenvector, diagonalization,...) has many applications in Science and Engineering.

69 A building is a coupled oscillator

70 A building is a coupled oscillator

71 A building is a coupled oscillator

72 A building is a coupled oscillator

73 A building is a coupled oscillator You need to diagonalize a large matrix to find what are the normal modes of the building!

74 Shinjyuku buildings at Earth Quake