Topic 1 shm

Oscillation and waves lecture notes

1. Topic 1: Simple Harmonic Motion (SHM)

2. Simple pendulum Flat disc SHM Systems

3. Mass-Spring Mass at the centre of a light string SHM Systems

4. Frictionless U-tube of liquid Open flask of volume V and neck of length l SHM Systems

5. Hydrometer LC circuit SHM Systems

6. When the above systems are slightly disturbed from the equilibrium or rest position , they will oscillate with SHM A small displacement x from its equilibrium position sets up a restoring force F , which is proportional to x acting in a direction towards the equilibrium position F = – sx (Hooke’s Law of Elasticity) s = proportional constant (called the stiffness) negative sign shows that the force is acting against the direction of increasing displacement and back towards the equilibrium position

7. By Newton’s Law, Dimensions of T = period of oscillation f = frequency of oscillation

8. Since the behaviors of x with time has a sinusoidal dependence, so it more appropriate to consider the angular frequency  = 2 f

9. Displacement in Simple Harmonic Motion (SHM) The behaviour of SHM is expressed in terms of its: : displacement x from equilibrium : velocity : acceleration at any given time A = constant with the same dimensions as x One possible solution: satisfies

10. Displacement in Simple Harmonic Motion (SHM) another solution: Complete or general solution for

11. Displacement in Simple Harmonic Motion (SHM) Rewrite: where  is a constant angle where a is the amplitude of the displacement and  is the phase constant Limiting values of sin(t+) are 1, so the system will oscillate between the values of x =  a

12. Displacement in Simple Harmonic Motion (SHM) Displacement x can be represented by projection of a radius vector of constant length a rotates in anticlockwise direction with constant angular velocity  Phase constant  defines the position in the cycle of oscillation at time t = 0  takes the values between 0 and 2 radian

13. Sinusoidal displacement of simple harmonic oscillator with time, showing v ariation of starting point in cycle in terms of phase angle  [source: H.J.Pain , The Physics of Vibrations and Waves 6 th Ed. Fig. 1.2] Displacement In Simple Harmonic Motion (SHM)

14. Velocity and Acceleration in SHM Displacement Velocity Acceleration Maximum value of velocity a  ( velocity amplitude) Maximum value of acceleration a  2 (acceleration amplitude)

15. Velocity and Acceleration in SHM Velocity leads the displacement by /2 Velocity is maximum when displacement is zero Velocity is zero when displacement is maximum Acceleration is anti-phase ( rad ) with respect to displacement Acceleration is maximum positive when displacement is maximum negative and vice versa

16. Variation with time of displacement, velocity and acceleration in SHM [source: H.J.Pain , The Physics of Vibrations and Waves6 th Ed. Fig. 1.3]

17. Two oscillators having the same frequency and amplitude may be consider in terms of their phase difference  1 -  2 When two systems are diametrically opposed, the system are anti-phase: phase difference  1 -  2 = n  rad ( n is odd integer) When two systems are exactly equal values of displacement, velocity and acceleration, the system are in phase : phase difference  1 -  2 = 2 n  rad ( n is any integer) Velocity and Acceleration in SHM

18. Blue curve: Red curve: Differ in amplitude only

19. Blue curve: Red curve: Differ in period only T  = T /2   = 2 

20. Blue curve: Red curve: Differ in phase only

21. Example 1.1 What are the angular frequency, the frequency, the period , the amplitude of the resulting motion? What is the amplitude of the oscillation? What is the maximum speed of the oscillating block, and where is the block when it occurs? What is the amplitude of the maximum acceleration of the block? What is the phase constant for the motion? What is the displacement function x ( t ) for the spring-block system? A block whose mass m is 680 g is fastened to a spring whose spring constant k is 65 N/m. The block is pulled a distance x = 11 cm from its equilibrium position at x = 0 on a frictionless surface and released from rest at t = 0.

22. Example 1.1 The block-spring system The block moves in SHM once it has been pulled to the side and released

23. Example 1.1 - Solution What are the angular frequency, the frequency, the period of the resulting motion? The block-spring system forms a linear SHM angular frequency: frequency: period:

24. (b) What is the amplitude of the oscillation? Example 1.1 - Solution The block is pulled a distance x = 11 cm from its equilibrium position at x = 0 on a frictionless surface and released from rest at t = 0  The amplitude x m = 11 cm

25. Example 1.1 - Solution (c) What is the maximum speed of the oscillating block, and where is the block when it occurs? The maximum speed occurs when the oscillating block is moving through the origin, i.e. at x = 0 Maximum speed = velocity amplitude v m = x m   v m = x m  = (0.11 m) (9.78 rad/s) = 1.1 m/s

26. Example 1.1 - Solution (d) What is the amplitude of the maximum acceleration of the block? amplitude of maximum acceleration = x m  2  a m = x m  2 = (0.11 m) (9.78 rad/s) 2 = 11 m/s The maximum acceleration occurs when the block is at the ends of its path

27. Example 1.1 - Solution (e) What is the phase constant for the motion? At time t = 0, the block is located at x = x m

28. (f) What is the displacement function x ( t ) for the spring-block system? Example 1.1 - Solution

29. Energy of a SHM An exchange between kinetic and potential energy In an ideal case the total energy remains constant but this is never realized in practice If no energy is dissipated: 1. E total = KE + PE = KE max = PE max 2. Amplitude a remains constant When energy is lost the amplitude gradually decays

30. Energy of a SHM Potential energy PE is found by summing all the small elements of work sx  dx (force sx times distance dx ) done by the system against the restoring force over the range 0 to x where x = 0 gives zero potential energy Kinetic energy

31. Energy of a SHM Total Energy: Since total energy E is constant

32. Energy of a SHM Maximum potential energy occurs at therefore Maximum kinetic energy is

33. Energy of a SHM The total energy at any instant of time t or position of x is: slide-9 = PE max

34. Energy At x = 0 the energy is all kinetic At x =  a the energy is all potential

35. Potential energy PE ( t ), kinetic energy K ( t ) and total energy E as function of time t for a linear harmonic oscillator

36. A 0.5 kg cart connected to a light spring for which the force constant is 20 N m -1 oscillates on a horizontal, frictionless air track. (a) Calculate the total energy of the system and the maximum speed of the cart if the amplitude of the motion is 3 cm. (b) What is the velocity of the cart when the position is 2 cm. (c) Compute the kinetic and potential energy of the system when the position is 2.00 cm. Example 1.2

37. Example 1.2 - Solutions Total energy = Maximum speed (b) (c) Kinetic energy = Potential energy =

38. SHM in an Electrical System An inductor L is connected across the plates of a capacitor C The force equation becomes the voltage equation q/C _ + _ + I

39. SHM in an Electrical System In the absence of resistance, the energy of the electrical system remains constant Exchanged between magnetic field energy stored in the inductor and electric field energy stored between the plates of the capacitor The voltage across the inductor is where I is the current flowing and q is the charge on the capacitor

40. SHM in an Electrical System The voltage across the capacitor is q/C Kirchhoff’s law

41. Energy stored in the magnetic field Electrical energy stored in the capacitor SHM in an Electrical System

42. Force or Voltage Energy Mechanical Force Electrical Voltage Comparison between Mechanical and Electrical Oscillators

43. Superposition of Two Simple Harmonic Vibrations in One Dimension Case-1: Vibrations having equal frequencies Displacement of first motion: where  =  2 -  1 is constant Resulting displacement amplitude R: Displacement of second:

44. Superposition of Two Simple Harmonic Vibrations in One Dimension Case-1: Vibrations having equal frequencies each representing SHM along the x axis at angular frequency  to give a resulting SHM displacement : x = R cos (  t + ) --- here shown for t = 0 Addition of Vectors

45. The phase constant  of R is Resulting simple harmonic motion has displacement An oscillation of the same frequency  but having an amplitude R and a phase constant  Superposition of Two Simple Harmonic Vibrations in One Dimension Case-1: Vibrations having equal frequencies

46. Displacement of first motion: Displacement of second: where  2 >  1 Resulting displacement: Superposition of Two Simple Harmonic Vibrations in One Dimension Case-2: Vibrations having different frequencies

47. Resulting oscillation at the average frequency (  2 +  1 )/2 Resulting amplitude of 2 a which modulates Amplitude varies between 2 a and zero under the influence of the cosine term of a much lower frequency (  2 -  1 )/2 Acoustically this growth and decay of the amplitude is registered as ‘beat’ of strong reinforcement when two sounds of almost equal frequency are heard The frequency of the ‘beat’ is (  2 -  1 ) Superposition of Two Simple Harmonic Vibrations in One Dimension Case-2: Vibrations having different frequencies

48. One along the x -axis, the other along the perpendicular y -axis Displacement, Eliminating t , get a general equation for an ellipse Superposition of Two Perpendicular Simple Harmonic Vibrations Having Equal Frequency

49. Expanding the arguments of the sines:  

50. Some special case: An ellipse: If a 1 = a 2 = a , a circle x 2 + y 2 = a 2 A straight line: A straight line: Superposition of Two Perpendicular Simple Harmonic Vibrations Having Equal Frequency

51. Superposition of Two Perpendicular Simple Harmonic Vibrations Having Equal Frequency

52. If the amplitudes of vibrations are a and b respectively, the Lissajous figure will always be contained within the rectangle of sides 2 a and 2 b The axial frequencies bear the simple ratio Superposition of Two Perpendicular Simple Harmonic Vibrations Having Different Frequency When the frequencies of the two perpendicular simple harmonic vibrations are not equal, the resulting motion becomes more complicated The patterns which are traced are called Lissajous figures

53. Simple Lissajous figures produced by perpendicular simple harmonic motions of different angular frequencies

54. Superposition of a Large Number n of Simple Harmonic Vibrations n simple harmonic vibrations of equal amplitude a and equal successive phase difference  Resultant magnitude R and phase difference 

55. Vector superposition of a large number n of simple harmonic vibrations of equal amplitude a and equal successive phase difference  resultant amplitude: its phase difference with respect to the first component a cos  t:

56. Geometrically, each length: where r is the radius of the circle enclosing the (incomplete) polygon From the isosceles triangle OAC the magnitude of the resultant: and its phase angle is seen to be: In the isosceles triangle OAC: In the isosceles triangle OAB:

57. i.e. half the phase difference between the first and the last contributions The resultant: magnitude of the resultant is not constant but depends on the value of 

58. When n is very large and  is very small and the polygon becomes an arc of the circle centre O, of length na = A , with R as the chord Then: In this limit:

59. The pattern is symmetric about the value = 0 and is zero whenever sin  = 0 except at   0 i.e. when sin  /   1

60. (b) When = 0,  = 0 and the resultant of the n vectors is the straight line of length A (c) As increases A becomes the arc of a circle until at  = / 2 the first and last contributions are out of phase (2  = ) and the arc A has become a semicircle of which the diameter is the resultant R (d) A further increase in  increases  and curls the constant length A into the circumference of a circle (  =  ) with a zero resultant (e) At  = 3 / 2, the length A is now 3/2 times the circumference of a circle whose diameter is the amplitude of the first minimum.

61. Rotating-Vector Representation of SHM Source: A.P. French, “Vibrations and Waves” MIT Introduction Physics Series, CRC Press

62. Describing SHM by regarding it as the projection of uniform circular motion of a disk of radius R rotates about a vertical axis at the rate of angular frequency  rad/s Small block Shadow of the block P The shadow performs SHM with period 2 / and amplitude A along a horizontal line on the screen

63. SHM as the geometrical projection of uniform circular motion Horizontal axis Ox = the line along which actual oscillation takes places Rotating vector OP projected onto a diameter of the circle Instantaneous position of the point P is defined by the constant length A and the variable angle 

64. Take counterclockwise direction as positive Actual value of  :  =  t +   is the value of  at t = 0 Displacement x of the actual motion: x = A cos  = A cos ( t + ) Orthogonal oscillation along y axis perpendicular to the real physical axis O x of the actual motion: y = A sin ( t + ) -- physical unreal component of the motion

65. Cartesian and polar representations of a rotating vector vector OP has the plane polar coordinates ( r , ) Cartesian coordinates ( x , y ) x = r cos  y = r sin  Rotating Vectors and Complex Numbers

66. Representation of a Vector in the Complex Plane z = a + jb j 2 =  1 jz = ja + j 2 b jz = b + ja Multiplication of z by j is equivalent to a 90° rotation

67. Sinusoidal signals are characterised by their magnitude , their frequency and their phase If the frequency is fixed (e.g. at the frequency of the AC supply) and we are interested in only magnitude and phase In such cases we often use phasor diagrams which represent magnitude and phase within a single diagram Phasor Diagrams

68. Phasor diagram is a graphical method that helps in the understanding waves and oscillations, and also helps with calculations, such as wave addition imagine a rigid rotor or vector moving around in circles around the origin, as illustrated in the diagram 1 the projection or shadow of the tip of the vector moves back and forth exactly like the oscillation

69. diagram 1 – Phasor Diagram the projection or shadow of the tip of the vector moves back and forth exactly like the oscillation The projection that the vector makes on the horizontal or  x -axis: x  =  A  cos ( ωt  +  ϕ ) A phasor is a complex number that represents the amplitude and phase of a sinusoid Phasors provide a simple means of analyzing linear system excited by sinusoidal sources

70. Representations of Complex Numbers Rectangular form: z = x + jy  Re x Im y jy z r Real part of z Imaginary part of z

71. Complex number z can be represented in three ways: z = x + jy Rectangular form: z = re j  Exponential form: e j  = cos  + j sin  x = r cos  y = r sin  z = r Polar form: 

72. Some Basic Properties of Complex Numbers e ±j  = cos  ± j sin  Real part: cos  = Re ( e j  ) Imaginary part: sin  = Im ( e j  ) x ( t ) = X m cos ( t +  ) Example: x ( t ) = Re( X e j t )  where X is the phasor representation of the sinusoid v ( t ) Phasor is a complex representation of the magnitude and phase of a sinusoid To obtain the sinusoid corresponding to a given phasor X , multiply the phasor by the time factor e jwt and take the real part X = X m e j  = X m  = Re( X m e j ( t +  ) ) = Re( X m e j  e j t )

73. x ( t ) = Re( X e j t ) To obtain the sinusoid corresponding to a given phasor X , multiply the phasor by the time factor e j  t and take the real part x ( t ) = X m cos ( t +  ) = Re( X m e j ( t +  ) ) = Re( X m e j  e j t ) To get the phasor corresponding to a sinusoid, we first express the sinusoid in the cosine form so that the sinusoid can be written as the real part of a complex number Then we take out the time factor e jwt , and whatever is left is the phasor corresponding to the sinusoid

74. Arithmetic Operations Complex Conjugate z 1 = x 1 + jy 1 = r 1  1 z 2 = x 2 + jy 2 = r 2  2 Addition: Division: Reciprocal: Multiplication: Subtraction: Square Root: –  1 –  2  1 +  2 /2 –

75. X m  X m X m x ( t ) = Re( X e j t ) X e j t = X m e j ( t + ) Plot of the sinor on the complex plane As time increases, the sinor rotates on a circle of radius of X m at an angular velocity  in the counterclockwise direction x ( t ) is the projection of the sinor X e jwt on the real axis the value of sinor at time t = 0 is the phasor X of the sinusoid x ( t ) X = X m e j  = X m  

76. Phasor diagrams can be used to represent the addition of signals. This gives both the magnitude and phase of the resultant signal

77. Phasor diagrams can also be used to show the subtraction of signals