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Physics Beyond 2000 Chapter 5 Simple Harmonic Motion ons/SHM.htm.

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2 Physics Beyond 2000 Chapter 5 Simple Harmonic Motion ons/SHM.htm

3 Simple Harmonic Motion It is a particular kind of oscillation. Abbreviation is SHM. Some terms Amplitude, period, frequency and angular frequency.

4 Definition of SHM The motion of the particle whose acceleration a is always directed towards a fixed point and is directly proportional to the distance x of the particle form that point. where ωis a constant, the angular frequency.

5 Examples of SHM 0 x = 0 The particle is at the position x = 0. It has a velocity to the right. V

6 Examples of SHM a A A is the maximum distance, the amplitude. 0 x The particle moves to the right with retardation a. Note that x and a are in opposite directions. x =A

7 Examples of SHM a A A is the maximum distance, the amplitude. 0 x The particle moves back to the left with acceleration a. Note that x and a are still in opposite directions. x = 0

8 Examples of SHM a A A is the maximum distance, the amplitude. 0 x The particle moves to the left with retardation a. Note that both x and a change directions. They are still in opposite directions. x = -A

9 Examples of SHM a A A is the maximum distance, the amplitude. 0 x The particle moves to the right with acceleration a. Note that x and a are still in opposite directions. x = 0

10 Examples of SHM The position x = 0 is the equilibrium position. 0 x = 0 V

11 Examples of SHM The position x = 0 is the equilibrium position at which the net force is zero. In the oscillation, 0 x V a The negative sign indicates the direction of a is opposite to that of x.

12 Example 1 Is it a SHM? a = -16.x

13 Differential equation of SHM acceleration a of SHM The left hand sides show the accelerations in different mathematical forms.

14 The kinematics of SHM Displacement x of a SHM. A solution of the differential equation is x = A.sin(ωt + ψ) t is the time, ωis the angular frequency, A is the amplitude and ψis the initial phase.

15 The kinematics of SHM Displacement x If ψ=0 x = A.sin (ωt) 0 t A -A T 2T3T time Displacement x = A.sin(ωt + ψ) of a SHM. The displacement x of a particle performing SHM changes sinusoidally with time t. The period of the SHM is

16 Phase and Initial Phase x = A.sin(ωt + ψ) ωt + ψis called the phase. At t = 0, phase reduces to ψ. x = A.sin(ψ) ψis called the initial phase.

17 Phase x = A.sin(ωt + ψ) If ψ=0, x = A.sin(ωt ) If ψ= π/2, x = A.cos(ωt ) If ψ= π, x = -A.sin(ωt ) etc.

18 Initial Phase ψ The value of ψis determined by the initial position of x (at t = 0). i.e. how the motion is started.

19 ψ= 0 x = A.sin(ωt) At t = 0, x = 0. –The motion starts at x = 0. In the first T/4, x increases with time t and approaches the amplitude A. 0 x = 0 V At t = 0 a

20 ψ= 0 x = A.sin(ωt) At t = 0, x = 0. –The motion starts at x = 0. In the first T/4, x increases with time t and approaches the amplitude A. Displacement x 0 time t A -A T 2T3T

21 ψ= π/2 x = A.cos(ωt) At t = 0, x = A. –The motion starts at x = A. In the first T/4, x decreases with time t and approaches 0. 0 x = A v = 0 At t = 0 a

22 ψ= π/2 x = A.cos(ωt) At t = 0, x = A. –The motion starts at x = A. In the first T/4, x decreases with time t and approaches 0. Displacement x 0 time t A -A T 2T3T

23 ψ= π x = -A.sin(ωt) At t = 0, x = 0. –The motion starts at x = 0. In the first T/4, x decreases with time t and approaches -A. 0 x = 0 V At t = 0 a

24 ψ= π x = -A.sin(ωt) At t = 0, x = 0. The motion starts at x = 0. In the first T/4, x decreases with time t and approaches -A. Displacement x 0 time t A -A T 2T3T

25 ψ= 2π/3 x = -A.cos(ωt) At t = 0, x = -A. The motion starts at x = -A. In the first T/4, x increases with time t and approaches 0. 0 x = -A v=0 At t = 0 a

26 ψ= 2π/3 x = -A.cos(ωt) At t = 0, x = -A. The motion starts at x = -A. In the first T/4, x increases with time t and approaches 0. Displacement x 0 time t A -A T 2T3T

27 Angular frequency ω Period T = time for one complete oscillation. Frequency f = number of oscillations in one second. Angular frequency = 2 f

28 Angular frequency ω In a SHM, x = A.sin(ωt+ψ) After a time T, x must be the same again. A.sin(ωt+ψ) = A.sin(ωt+ ωT +ψ) ωT = 2π Unit of ω is rad s -1

29 Isochronous oscillation The period T in a SHM is independent of the amplitude A. A SHM is an isochronous oscillation.

30 Velocity in SHM x = A.sin(ωt+ψ)

31 Velocity in SHM x = A.sin(ωt+ψ) What is the maximum speed in this motion? v o = Aω The maximum speed occurs at the equilibrium position. i.e. when x = 0.

32 Velocity in SHM Example 2.

33 Acceleration in SHM

34 What is the maximum acceleration in this motion? a o = -Aω 2 or Aω 2 The maximum acceleration occurs at the positions with maximum displacement. i.e. when x = A or -A.

35 Displacement, Velocity and Acceleration in SHM x = A.sin(ωt + ψ) v = Aω.cos(ωt + ψ) a = -Aω 2.sin(ωt + ψ)

36 Displacement, Velocity and Acceleration in SHM with ψ=0 0 t A -A T 2T3T x v 0 t Aω -Aω T 2T3T a x = A.sin(ωt) v = Aω.cos(ωt) 0 t Aω 2 -Aω 2 T 2T3T a = -Aω 2.sin(ωt)

37 Acceleration and Displacement in SHM x = A.sin(ωt + ψ) a = -Aω 2.sin(ωt + ψ) a = - ω 2.x This is a characteristic of a SHM.

38 Acceleration and Displacement in SHM a = - ω 2.x acceleration and displacement in SHM are in antiphase.(i.e. in opposite directions.) 0 t A -A T 2T3T x = A.sin(ωt) x 0 t Aω 2 -Aω 2 T 2T3T a = -Aω 2.sin(ωt) a

39 Velocity and Displacement in SHM 0 t A -A T 2T3T x = A.sin(ωt) x v 0 t Aω -Aω T 2T3T v = Aω.cos(ωt) v leads x by π/2 or 90 0 (or x lags behind x by π/2 or 90 0 ).

40 Velocity and Displacement in SHM v leads x by π/2 or 90 0.

41 Velocity and Acceleration in SHM v 0 t Aω -Aω T 2T3T v = Aω.cos(ωt) a leads v by π/2 or 90 0 (or v lags behind a by π/2 or 90 0 ). 0 t Aω 2 -Aω 2 T 2T3T a = -Aω 2.sin(ωt) a

42 The kinematics of SHM We may look upon SHM as a projection of a uniform circular motion on its diameter. A green ball is performing a uniform circular motion with angular velocity. ω

43 The kinematics of SHM A green ball is performing a uniform circular motion with angular velocity. Its projection on the diameter is the red ball performing SHM. ω on/vrmlshm.html

44 The kinematics of SHM A green ball is performing a uniform circular motion with angular velocity. Its projection on the diameter is the red ball performing SHM. ω

45 The kinematics of SHM A green ball is performing a uniform circular motion with angular velocity. Its projection on the diameter is the red ball performing SHM. ω

46 The kinematics of SHM A green ball is performing a uniform circular motion with angular velocity. Its projection on the diameter is the red ball performing SHM. ω

47 The kinematics of SHM A green ball is performing a uniform circular motion with angular velocity. Its projection on the diameter is the red ball performing SHM. ω

48 Period The period of the circular motion is the same as the period of the SHM. Period T = time to complete one revolution Period T = time to move to and fro once.

49 Displacement of SHM Let A = radius of the circle It is also the amplitude of the SHM Let be the starting angle of the green ball. It is the initial phase of the red ball. At time = 0, the position of the green ball and the red ball are as shown. ω A O

50 Displacement of SHM A O After time = t, the green travels and angular displacement t and the red ball moves a displacement x as shown. ω x t The displacement of SHM is x = A.sin( t + ) t + is the phase of the SHM.

51 Velocity of SHM A + t O ω v The linear speed of the green ball is A. The velocity of the red ball is the horizontal component of A. v = A.cos( t + ) Let A = v o v = v o.cos( t + ) A

52 Acceleration of SHM A 2 + t O ω The green has centripetal acceleration A. 2 The acceleration of SHM is the horizontal component of A. 2 v a a = -A 2.sin( t + ) Let A 2 = a o a = - a o.sin( t + )

53 SHM and Uniform Circular Motion A 2 + t O ω v a x = A.sin( t + ) and a = -A 2.sin( t + ) a = - 2.x

54 SHM and Uniform Circular Motion A 2 + t O ω a The projection of a uniform circular motion on its diameter is SHM.

55 SHM and Uniform Circular Motion A 2 O ω A In a SHM, the red ball passes its equilibrium position: 1. The speed of the red ball is a maximum. 2. The acceleration of the red ball is zero.

56 SHM and Uniform Circular Motion A 2 O ω A In a SHM, when the red ball is at its maximum displacement: 1.The speed of the red ball is a zero. 2.The acceleration of the red ball is a maximum.

57 Rotating Vector It is difficult to represent SHM using diagram because its quantities vary with time. Use a vector which is rotating in uniform circular motion. urendranath/Shm/Shm01.ht mlhttp://members.nbci.com/S urendranath/Shm/Shm01.ht ml O

58 Rotating Vector It is difficult to represent SHM using diagram because its quantities vary with time. Use a vector which is rotating in uniform circular motion. O

59 Rotating Vector It is difficult to represent SHM using diagram because its quantities vary with time. Use a vector which is rotating in uniform circular motion. O

60 Rotating Vector It is difficult to represent SHM using diagram because its quantities vary with time. Use a vector which is rotating in uniform circular motion. O

61 Rotating Vector It is difficult to represent SHM using diagram because its quantities vary with time. Use a vector which is rotating in uniform circular motion. O

62 Rotating Vector Use a vector which is rotating in uniform circular motion. Its projection on the diameter is SHM. Period T = O

63 Rotating Vector Use three rotating vectors. The three projections on the diameter represent displacement, velocity and acceleration of the SHM respectively. The magnitude of the displacement vector is x o. The magnitude of the velocity vector is v o. The magnitude of the acceleration vector is a o. O xoxo vovo aoao

64 Rotating Vector O xoxo vovo aoao From the rotating vectors, the velocity v leads displacement x by π/2 and acceleration a and displacement x are in antiphase. modules/m9/oscillations.htm

65 The Dynamics of SHM a x v When the particle is at a displacement x, there is an acceleration a. From Newtons 2 nd law, there must be a net force F net on it F net = m.a and in SHM a = -ω 2.x The above two equations F net = -mω 2.x

66 Horizontal Block-spring system Equilibrium position Is this a SHM? The block is oscillating. Mass of the block = m ; Mass of the spring can be ignored. There is not any friction.

67 Horizontal Block-spring system Equilibrium position The block is oscillating. Suppose that the block has been displaced by a displacement x. F net = m.a and F net = -k.x where k is the spring constant x m.a = -k.x a = -.x

68 Horizontal Block-spring system Equilibrium position The block is oscillating. x a = -.x Compare it to a standard SHM equation a = -.x We see that the motion of the block is a SHM with ω = a

69 Horizontal Block-spring system Equilibrium position The block is oscillating. x a Its period T = The period T is independent of the amplitude of oscillation. This is an isochronous oscillation.

70 Horizontal Block-spring system Equilibrium position The force is maximum when the displacement is a maximum. The force is zero at the equilibrium position. F net

71 Horizontal Block-spring system Equilibrium position The speed is zero when the displacement is a maximum. The speed is fastest at the equilibrium position. v

72 Horizontal Block-spring system A simulation program You may download the program from –http://www.programfiles.com/index.asp?ID=74 91http://www.programfiles.com/index.asp?ID=74 91

73 Vertical Block-spring system /explrsci/dswmedia/harmonic.htmhttp://webphysics.ph.msstate.edu/javamirror /explrsci/dswmedia/harmonic.htm ena/simple_harmonic_motion.htmlhttp://www.exploratorium.edu/xref/phenom ena/simple_harmonic_motion.html astr.gsu.edu/hbase/shm.htmlhttp://hyperphysics.phy- astr.gsu.edu/hbase/shm.html

74 Vertical Block-spring system natural length Equilibrium position e = extension mg ke m = mass of the block k = spring constant mg = ke Mass of the spring can be ignored. oscillation Is this a SHM?

75 Vertical Block-spring system oscillation natural length Equilibrium position Suppose that the block has moved a displacement x below the equilibrium position. At the instant, what are the forces acting on the block? x e

76 Vertical Block-spring system oscillation natural length Equilibrium position F net = -k(x+e)+mg and mg = ke F net = -k.x x k(x+e) e mg

77 Vertical Block-spring system oscillation natural length Equilibrium position F net = -k.x and F net = m.a m.a = -k.x x k(x+e) e mg a = -.x

78 Vertical Block-spring system oscillation natural length Equilibrium position x k(x+e) e mg a = -.x So it is a SHM with and Its period T =

79 Vertical Block-spring system The motion is similar to that of a horizontal block spring system. At the equilibrium position, the net force on the block is zero though the tension of the spring is ke. At the equilibrium position, the speed of the block is fastest. At the maximum displacement, the net force is a maximum and the speed is zero.

80 Examples Example 4 Horizontal block-spring system. Example 5 Vertical block-spring system. Example 6 Combination of springs.

81 Simple pendulum The simple pendulum is set into oscillation. Is it a SHM? r/explrsci/dswmedia/harmonic.htm

82 Simple pendulum Equibrium position θ is the length of the simple pendulum. m is the mass of the bob. Suppose that the bob has an angular displacement θ from the equilibrium position.

83 Simple pendulum Equibrium position θ What are the forces acting at the bob at this instant?

84 Simple pendulum What are the forces acting at the bob at this instant? T = tension from the string mg = weight of the bob Equilibrium position θ T mg

85 Simple pendulum As the bob is in a circular motion, there is a net force (the centripetal force) on the bob. Equilibrium position θ T mg

86 Simple pendulum The tangential component F t = -mg.sin For small angle, sin. So F t -mg. ……… (1) Only the tangential component F t is involved in the change of speed of the bob. So F t = m.a …………. (2) Equilibrium position θ T FtFt FrFr

87 Simple pendulum Equilibrium position θ T FtFt FrFr x From equations (1) and (2), a = -g. ………….. (3) The displacement of the bob is x …………. (4)

88 Simple pendulum Equilibrium position θ T FtFt FrFr x From equations (3) and (4), a = -.x So it is a SHM with Note that this is true for small angle of oscillation

89 Simple pendulum Equilibrium position θ T FtFt FrFr x Note that this is true for small angle of oscillation The period of oscillation is

90 Example 7 To find the period of a simple pendulum

91 A floating object It is a SHM. water mg Floating force In equilibrium

92 A floating object It is a SHM. water

93 A floating object It is a SHM. water

94 Liquid in a U-tube It is SHM.

95 Energy in SHM There is a continuous interchange of kinetic energy and potential energy. Horizontal Block-spring system –kinetic energy of the block elastic potential energy of the block Simple pendulum –kinetic energy of the bob gravitational potential energy

96 Energy in SHM Kinetic energy is maximum when the particle passes the equilibrium position. The maximum kinetic energy is

97 Energy in SHM Kinetic energy is maximum when the particle passes the equilibrium position. If we write

98 Energy in SHM Kinetic energy is zero when the particle reaches its maximum displacement A. Kinetic energy is completely transformed into potential energy. The maximum potential energy U po is equal to the maximum kinetic energy U ko

99 Energy in SHM If the energy is conserved (the motion is undamped), the total energy is U o = U ko = U po At any position, the total energy is

100 Example 10 Angular speed of a simple pendulum

101 Energy versus displacement When the displacement = x, –the potential energy U p = k.x 2 –the kinetic energy U k = m.v 2 –the total energy U o = k.A 2 –U p + U k = U o

102 Energy versus displacement 0A-A x Energy UkUk UpUp UoUo

103 Example 11 Find the total energy and the maximum speed.

104 Energy versus time For simplicity, choose ψ= 0. Use x = A.sin(ωt) and v = A ω.cos(ωt)

105 Energy versus time For simplicity, choose ψ= 0. Use x = A.sin(ωt) and v = A ω.cos(ωt)

106 Energy versus time time 0T2T Displacement UpUp UkUk

107 Damped Oscillation There is loss of energy in this kind of oscillation due to friction or resistance. As a result, the amplitude decreases with time. There are –Slightly damping –Critical damping –Heavy damping fun/JAVA/dho/dho.html olgar/java/Hooke/Hooke.html

108 Slightly damping The amplitude decreases exponentially with time. A n = A 1. n where is a constant

109 Critical damping The system does not oscillate but comes to rest at the shortest possible time. –e.g. galvanometer Displacement x time t 0 T2T

110 Heavy damping The resistive force is very large. The system returns very slowly to the equilibrium position. Displacement x time t 0 T2T

111 Example 12 Slightly damping

112 Forced oscillations The vibrating system is acted upon by an external periodic driving force. The system is vibrating at the frequency of the external force. Without the external driving force, the frequency of free oscillation is called natural frequency. With the external periodic driving force, the amplitude of vibration varies ts/a-city/physengl/resonance.htmhttp://physics.uwstout.edu/physapple ts/a-city/physengl/resonance.htm hand vibrating up and down

113 Forced oscillations The amplitude of a forced oscillation depends on –damping –frequency of the driving force hand vibrating up and down

114 Forced oscillations The amplitude of a forced oscillation is a maximum when the frequency of the driving force is equal to the natural frequency f o of the system hand vibrating up and down fofo Amplitude Driving frequency no damping slight damping heavy damping

115 Forced oscillations Resonance occurs when the driving frequency equals the natural frequency. The amplitude is a maximum at resonance. hand vibrating up and down fofo Amplitude Driving frequency no damping slight damping heavy damping

116 Phase relationship When the driving frequency is less than the natural frequency f o of the vibrating system, the two motions are in phase. When the driving frequency is greater than or equal to the natural frequency f o of the vibrating system, the driven motion always lags behind the driving motion.

117 Phase relationship Driving frequency Driven motion Phase lag Less than f o 0 equal to f o /2 greater than f o slight damping heavy damping /2 0 fofo driving frequency

118 Tacoma Narrows Bridge The bridge was driven by the wind.

119 Tacoma Narrows Bridge The bridge was in resonance with the wind.

120 Tacoma Narrows Bridge The bridge was in resonance with the wind and finally collapsed.

121 Why is a simple harmonic motion simple?


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