2. 2
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Simple pendulum Flat disc
SHM Systems
3. 3
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Mass-Spring Mass at the centre of a light string
SHM Systems
4. 4
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Frictionless U-tube of liquid
Open flask of volume V and
neck of length l
SHM Systems
5. 5
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Hydrometer LC circuit
SHM Systems
6. 6
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
• 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. 7
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
• By Newton’s Law, xmF
sxxm
0 sxxm
2
2
dt
xd
x
0 x
m
s
x
22
2
fT
ML
MLT
m
s
• Dimensions of
• T = period of oscillation
• f = frequency of oscillation
8. 8
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
• Since the behaviors of x with time has a sinusoidal dependence,
so it more appropriate to consider the angular frequency = 2f
2
2
1
s
m
f
T
m
s
2
02
xx
9. 9
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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
x
x
xtAx
tAx
tAx
22
cos
sin
cos
A = constant with the same dimensions as x
• One possible solution: satisfiestAx cos 02
xx
10. 10
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Displacement in Simple Harmonic Motion (SHM)
xtBx
tBx
tBx
22
sin
cos
sin
• another solution: tBx sin
• Complete or general solution for
xtBtAx
tBtAx
22
)sincos(
sincos
02
xx
11. 11
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Displacement in Simple Harmonic Motion (SHM)
• Rewrite:
where is a constant angle
sinaA cosaB
22
222222
)cos(sin
BAa
aaBA
)sin(
sincoscossin
tax
tatax
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. 12
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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. 13
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Sinusoidal displacement of simple harmonic oscillator with time,
showing variation of starting point in cycle in terms of phase angle
[source: H.J.Pain, The Physics of Vibrations and Waves 6th Ed. Fig. 1.2]
Displacement In Simple Harmonic Motion (SHM)
14. 14
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Velocity and Acceleration in SHM
• Displacement
• Velocity
• Acceleration
• Maximum value of velocity a (velocity amplitude)
• Maximum value of acceleration a2 (acceleration amplitude)
)sin( tax
)cos( tax
dt
dx
)sin(2
2
2
tax
dt
xd
15. 15
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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. 16
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Variation with time of displacement, velocity and acceleration in SHM
[source: H.J.Pain, The Physics of Vibrations and Waves6th Ed. Fig. 1.3]
17. 17
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
• 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 = 2n rad ( n is any integer)
Velocity and Acceleration in SHM
19. 19
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
txx m cosBlue curve:
txx m cos'
Red curve:
Differ in amplitude only
20. 20
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
txx m cosBlue curve:
txx m 'cosRed curve:
Differ in period only
T = T/2
= 2
21. 21
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
txx m cosBlue curve:
)4/cos( txx mRed curve:
Differ in phase only
22. 22
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Example 1.1
a) What are the angular frequency, the frequency, the period , the
amplitude of the resulting motion?
b) What is the amplitude of the oscillation?
c) What is the maximum speed of the oscillating block, and
where is the block when it occurs?
d) What is the amplitude of the maximum acceleration of the
block?
e) What is the phase constant for the motion?
f) 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.
23. 23
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Example 1.1
The block-spring system
The block moves in SHM once it has been
pulled to the side and released
24. 24
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Example 1.1 - Solution
(a) What are the angular frequency, the frequency, the period of the
resulting motion?
The block-spring system forms a linear SHM
angular frequency: rad/s78.9
kg0.68
N/m65
m
k
frequency: Hz56.1
2
rad/s78.9
2
f
period: s64.0
1
f
T
25. 25
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
(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 xm = 11 cm
26. 26
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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 vm = xm
vm = xm = (0.11 m) (9.78 rad/s) = 1.1 m/s
27. 27
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Example 1.1 - Solution
(d) What is the amplitude of the maximum acceleration of the
block?
amplitude of maximum acceleration = xm 2
am = xm 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
28. 28
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Example 1.1 - Solution
(e) What is the phase constant for the motion?
At time t = 0, the block is located at x = xm
)cos( txx m
0
1cos)0cos(
mm xx
29. 29
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
(f) What is the displacement function x(t) for the spring-block
system?
)8.9cos(.11.0)(
])0)rad/s8.9cos([()m11.0()(
ttx
ttx
)cos()( txtx m
Example 1.1 - Solution
30. 30
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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. Etotal = KE + PE = KEmax = PEmax
2. Amplitude a remains constant
• When energy is lost the amplitude gradually decays
31. 31
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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
2
0 2
1
sxdxsxPE
x
• Kinetic energy 2
2
1
xmKE
32. 32
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Energy of a SHM
• Total Energy:
22
2
1
2
1
sxxmE
0)( xsxxm
dt
dE
• Since total energy E is constant
0 sxxm
PEKEE
33. 33
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Energy of a SHM
• Maximum potential energy occurs at thereforeax
2
max
2
1
saPE
22
max
max
222
max
2
max
2
1
)(cos
2
1
2
1
maKE
tma
xmKE
• Maximum kinetic energy is
34. 34
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Energy of a SHM
The total energy at any instant of time t or position of x is:
2
22
2222
22
2
1
2
1
)(sin)(cos
2
1
2
1
2
1
saE
ma
ttma
sxxmE
)cos( tax
dt
dx
)sin( tax
2
ms
slide-9
= PEmax
35. 35
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Energy
At x = 0
the energy is all kinetic
At x = a
the energy is all potential
36. 36
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Potential energy PE(t), kinetic energy K(t) and total energy E as
function of time t for a linear harmonic oscillator
37. 37
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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
38. 38
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Example 1.2 - Solutions
(a) Total energy = J009.003.020
2
1
2
1 22
kA
1-2
sm19.0009.0)5.0(
2
1
xx Maximum speed
1-
22
sm141.0
009.002.020
2
1
5.0
2
1
x
x
(b)
J005.0141.05.05.0 2
J004.002.0205.0 2
(c) Kinetic energy =
Potential energy =
39. 39
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
SHM in an Electrical System
• An inductor L is connected
across the plates of a capacitor
C
• The force equation becomes the
voltage equation
dt
dI
L q/C
_
+
_
+
I
0
C
q
qL
40. 40
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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
2
2
dt
qd
L
dt
dI
LV
41. 41
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
SHM in an Electrical System
• The voltage across the capacitor is q/C
• Kirchhoff’s law
LC
qq
C
q
qL
1
0
0
2
2
42. 42
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
• Energy stored in the magnetic field
• Electrical energy stored in the capacitor
22
0 2
1
2
1
qLLILIdIE
Idt
dt
dI
LdtVIE
I
L
L
C
q
CV
22
1 2
2
SHM in an Electrical System
2
q
43. 43
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Force or Voltage Energy
Mechanical
Force
Electrical
Voltage
0sxxm
0
C
q
qL
22
2
1
2
1
sxxmE
C
q
qLE
2
2
2
1
2
1
Comparison between
Mechanical and Electrical Oscillators
44. 44
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Superposition of Two Simple Harmonic
Vibrations in One Dimension
Case-1: Vibrations having equal frequencies
)cos( 111 tax
)cos( 222 tax
cos2
)sin()cos(
21
2
2
2
1
2
2
2
2
21
2
aaaaR
aaaR
Displacement of first motion:
where = 2 - 1 is constant
Resulting displacement amplitude R:
Displacement of second:
45. 45
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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
46. 46
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
• 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
2211
2211
coscos
sinsin
tan
aa
aa
)cos( tRx
Superposition of Two Simple Harmonic
Vibrations in One Dimension
Case-1: Vibrations having equal frequencies
47. 47
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Displacement of first motion:
Displacement of second:
where 2 > 1
Resulting displacement:
tax 11 sin
tax 22 sin
2
)(
cos
2
)(
sin2
)sin(sin
1221
2121
tt
ax
ttaxxx
Superposition of Two Simple Harmonic
Vibrations in One Dimension
Case-2: Vibrations having different frequencies
49. 49
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
• Resulting oscillation at the average frequency (2+ 1)/2
• Resulting amplitude of 2a which modulates
• Amplitude varies between 2a 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
50. 50
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
• One along the x-axis, the other along the perpendicular y-axis
• Displacement,
• Eliminating t, get a general equation for an ellipse
)sin( 11 tax
)sin( 22 tay
)(sin)cos(
2
12
2
12
21
2
2
2
2
1
2
aa
xy
a
y
a
x
Superposition of Two Perpendicular
Simple Harmonic Vibrations Having Equal Frequency
51. 51
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Expanding the arguments of the sines:
52. 52
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
• Some special case:
• An ellipse:
If a1 = a2 = a, a circle x2 + y2 = a2
• A straight line:
• A straight line:
2
12
12
2
2
2
1
2
a
y
a
x
...,4,2,012 x
a
a
y
1
2
...,5,3,12 x
a
a
y
1
2
Superposition of Two Perpendicular
Simple Harmonic Vibrations Having Equal Frequency
53. 53
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Superposition of Two Perpendicular
Simple Harmonic Vibrations Having Equal Frequency
54. 54
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
• If the amplitudes of vibrations are a and b respectively, the
Lissajous figure will always be contained within the rectangle of
sides 2a and 2b
• 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
55. 55
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Simple Lissajous figures produced by perpendicular simple
harmonic motions of different angular frequencies
56. 56
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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
57. 57
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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 acos t:
58. 58
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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:
59. 59
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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
60. 60
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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:
61. 61
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
The pattern is symmetric about the
value = 0 and is zero whenever
sin = 0 except at
0 i.e. when sin / 1
62. 62
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
(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.
63. 63
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Rotating-Vector Representation
of SHM
Source: A.P. French, “Vibrations and Waves” MIT
Introduction Physics Series, CRC Press
64. 64
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
• 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
65. 65
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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
66. 66
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
• 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 Ox of the actual motion:
y = A sin (t + )
-- physical unreal component of the motion
67. 67
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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
68. 68
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Representation of a Vector in the Complex Plane
z = a + jb
j 2 = 1
jz = ja + j2b
jz = b + ja
Multiplication of z by j is
equivalent to a 90° rotation
69. 69
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
• 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
70. 70
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
• 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
71. 71
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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
72. 72
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Representations of Complex Numbers
Rectangular form: z = x + jy
Re
x
Im
y
jy
z
r
Real part of z
)Re(zx
Imaginary part of z
)Im(zy
1j
12
j
123
jjj
73. 73
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
• Complex number z can be represented in three ways:
z = x + jyRectangular form:
z = re jExponential form:
e j = cos + j sin
22
yxr
x
y1
tan
x = r cos y = r sin
1j
z = rPolar form:
j
rerjrrjyxz sincos
74. 74
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
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) = Xm cos (t + )
Example:
x(t) = Re(X e jt
)
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 = Xm e j
= Xm
= Re(Xm e j(t + )
) = Re(Xm e j
e jt
)
75. 75
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
x(t) = Re(X e jt
)
To obtain the sinusoid corresponding to
a given phasor X, multiply the phasor
by the time factor e jt and take the real
part
x(t) = Xm cos (t + ) = Re(Xm e j(t + )
) = Re(Xm e j
e jt
)
• 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
77. 77
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
Xm
Xm
Xm
x(t) = Re(X e jt
)
Xe jt
= Xm e j(t +)
Plot of the sinor on the complex plane
As time increases, the sinor rotates
on a circle of radius of Xm at an
angular velocity in the
counterclockwise direction
x(t) is the projection of the sinor
Xejwt on the real axis
the value of sinor at time t = 0 is the
phasor X of the sinusoid x(t)
X = Xm e j
= Xm
78. 78
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
• Phasor diagrams can be used to represent the addition of
signals. This gives both the magnitude and phase of the
resultant signal
79. 79
Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves
• Phasor diagrams can also be used to show the subtraction of
signals