al velocity and wave propagation velocity.
Undisturbed
position of string
ring is moving to the right with a
p and down in SHM about the
he string.
vy changes from time to time
50
Particle vibrational velocity Th
(or
Velocity of particle ( ) which oscillate about as
equilibrium position in simple harmonic motion. Fig
rig
By differentiating the displacement equation of the
Th
wave, thus ; eq
vy = dy = d [ A sin (ω t ± k x)]
dt dt
vy = Aω cos (ω t ± k x)
Particle vibrational velocity, vy is not constant and varies Th
with time, t .
Wave propagation velocity
he wave velocity ( ) is the velocity at which wave crests
r other part of waveform) move forward. It can be defined
s the distance travelled by a wave profile per unit time.
gure below shows a progressive wave profile moving to the
ght.
v
λ
he trough travels a distance of one wavelength, in a time
qual to one period, T. Thus, the wave velocity is:
v = distance v = λ where T = 1
time T f
∴ =
he wave velocity , vwave is constant
51
7.4 (e) Use particle vibrational velocity, vy = A
(f) Use wave propagation velocity, =
EXAMPLE 7.4.4 :
A progressive wave equation is given as : y = 3sin (8π
Find : (b) f
(a) Particle velocity, vy
(b) Velocity of the wave v
Solution: f
(a) vy = dy = Aω cos (ω t + k x)
dt
= (3) (8π ) cos (8π t + 0.4 x)
= 24π cos (8π t + 0.4 x) cm s−1 ∴
Aω cos (ω t ± k x)
π t + 0.4 x) cm
from : k = 2π = 2π = 5π
λ
0.4
λ = 2π
k
from :ω = 2π f
f = ω = 8π = 4 Hz
2π 2π
∴ wave velocity, v = f λ 52
= 4 (5π )
= 20π cm s−1
7.4 (g) Analyse the graphs of: (i) displacemen
(ii) displacemen
i) Graph of displacement against time, y-t
The graph shows the displacement of any one particle in the
For example, consider the equation of the wave is :
y = Asin (ωt − k
When x = 0 , thus
y = Asin (ωt − k(0))
y = Asin (ωt)
Hence, the displacement-time, y-t graph A
of = ( ) is :
0
−A
nt–time, y-t
nt–distance, y-x
e wave at any particular distance, x from the origin.
kx)
T T 3T 2T
22
53
ii) Graph of displacement against distance, y-x
The graph shows the displacement of all the particles in the wa
For example, consider the equation of the wave is :
y = Asin (ωt − k
When t = 0 , thus y
y = Asin (ω(0)− kx)
y = Asin (− kx)
y = −Asin (kx)
Hence, the displacement-distance, y-x graph A
of = − ( ) is :
0
−A
ave at any particular time, t.
kx)
v
x
λ λ 3λ 2λ
22
54
EXAMPLE 7.4.5 :
y (cm) 1.0
3
0
−3
Figure above shows a displacement, y against
progressive wave which propagates to the righ
a. Determine the wave number and frequency
b. Write the expression of displacement as a f
x (cm)
2.0
t distance, x graph after time, t for the
ht with a speed of 50 cm s−1.
y of the wave.
function of x and t for the wave above.
55
Solution: Given = 0.5 m s−1
(a) From the graph, = 1 = 1.0 × 10−2 m
= = 1.0 2
× 10−2
k = 200π m−1
by using the formula of wave speed, =
0.5 = 1.0 × 10−2
= Hz
(b) The expression is given by y(x, t ) = Asin (ωt − kx)
y(x,t) = 0.03sin (2π (50)t
y(x,t ) = 0.03sin (100πt −
Where y and x in metres and t i
− 200πx)
− 200πx)
in seconds
56
EXAMPLE 7.4.6 :
A progressive wave is represented by the equation , =
seconds.
a. Determine the angular frequency, the wavelength, the perio
b. Sketch the displacement against distance graph for progres
c. Sketch the displacement against time graph for the partic
d. Sketch the displacement against time graph for the parti
e. Is the wave traveling in the +x or –x direction?
f. What is the displacement y when t=5s and x=0.15cm.
Solution: Thus,
(a) By comparing the general equation
(i) ω = π rad s
, = −
(iii) The period :
with
= 2
, = −
= 2
T =2s
− where y and x are in centimeters and t in
od, the frequency and the wave speed.
ssive wave above in a range of 0 ≤ x ≤ λ at time, t = 0 s.
cle at x = 0 in a range of 0 ≤ t ≤ T.
icle at x = 0.5λ in a range of 0 ≤ t ≤ T.
(ii) k = π cm−1 and k = 2π 2π = π
λ λ
s −1
Therefore, = 2 cm
(iv) The frequency and wave speed:
= 1 = 1 f = 0.5 Hz
2
By applying the equation of wave speed v = λf
= 2 × 10−2 0.5
= 1 × 10−2 m s−1 57
(b) At time, t = 0 s, the equation of displacement as a func
, 0 = 2 sin 0 −
, 0 = 2 sin − y (cm)
, 0 = −2 sin
**Note: 2
Calculate λ first: 2 0
λ
since = and =
∴λ= 2 2 2
= =
−2
ction of distance, x is given by
, 0 = −2 sin
x (cm)
12
58
(c) The particle at distance, x = 0 , the equation of displace
0, = 2 sin − 0
0, = 2 sin
Hence the displacement, y against time, t graph is
y (cm) y(0,t ) = 2sin (πt )
2
t(
0 12
−2
ement as a function of time, t is given by
**Note:
Calculate first: 2
since = and =
2
∴ = 2 = 2 =
(s)
59
(d) The particle at distance, x = 0.5λ and λ = 2 cm thus x = 1 c
as a function of time, t is given by
y(1,t) = 2sin (πt −π (1))
y(1,t) = 2sin (πt −π )
Then the graph of displacement, y against time, t in the r
t 0 0.5 1 1.5 0 y (cm)
y 0 -2 0 2 0
2
0
−2
cm. Therefore, the equation of displacement
range of 0 ≤ t ≤ T is
t (s)
12
60
(e) +x direction
(f) y(x, t ) = 2 sin (πt − πx)
= 2 sin [π (5) − π (0.15)]
= 0.91cm
61
Exercise
A wave travelling along a string is described by
y(x,t) = 0.327sin (2.72t − 72.1x)
where y in cm, x in m and t is in seconds. Determin
a) the amplitude, wavelength and frequency of
wave.
b) the velocity with which the wave moves along
string.
c) the displacement of a particle located at x = 22
and t = 18.9 s.
ANS: 0.327 cm, 8.71 cm, 0.433 Hz; 0.0377 m s−1; −0.1
ne
the
g the
2.5 cm
192 cm
62
7.5 : Superposition of Wave
Learning Outcome:
a) State the principle of superposition
destructive interferences.
b) Use the standing wave equation =
c) Compare between progressive wav
es
n of waves for the constructive and
= 2 cos sin
ves and standing waves.
63
7.5 (a) State the principle of superposition of w
interferences.
Superposition of waves means when two wave
Principle of superposition of waves: At any t
interfer
of the i
Where y1
waves for the constructive and destructive
es meet in the same region, they interfere.
time, the combined waveform of two or more
ring waves is given by the sum of the displacements
individual waves at each point in the medium.
y = y1 + y2
1 and y2 are the displacement of individual pulses at that point.
64
Interference is defined as the interaction (superp
1. Constructive interference : Before
interference
the vertical displacements of the
two pulses are in the same During
direction, and the amplitude of interference
the combined waveform is greater
than that of either pulse.
• It occurs wave pulse 1 and 2 have the
same wavelength, frequency and in
phase each other.
After
interference
position) of two or more wave motions.
wave pulse 1 wave pulse 2
A
Wave pulse 1 and Wave pulse 2 are
A travelling against each other. Both
have the same amplitude, A.
e A+A The amplitude of the wave doubles
(A + A) when both waves interfered.
2A
Resultant displacement, y = 2A
Both waves travel with the initial
A A amplitude, A.
65
2. Destructive interference : wave
one pulse has a negative Before A
displacement, the two pulses tend interference
to cancel each other when they
overlap, and the amplitude of the During
combined waveform is smaller interference
than that of either pulse.
• It occurs wave pulse 1 and 2 have the
same wavelength, frequency and
antiphase each other.
After
interference
--A
pulse 1 Wave pulse 1has positive amplitude while
A wave pulse 2 has negative amplitude.
-A
wave pulse 2
A+(-A) the sum of the amplitude for wave A+(-A) is zero .
Resultant displacement, y = 0
Both waves propagate with initial
A amplitude.
A
66
7.5 (b) Use the standing wave equation =
• wave whose profile do not a
medium to the other end.
• Standing wave appear to be
• produced by the superpositi
frequency, traveling in oppo
• also known as stationary wa
advance & no energy is transferred from one end of the
e vibrating vertically without traveling horizontally.
ion of two progressive waves of equal amplitude and
osite direction.
aves.
67
By considering the wave functions for two progressive waves:
y (m)
+ 2A
standing wave:
− 2A
y1 = Asin(ωt − kx) and y2 = Asin(ωt + kx)
Applying principle of superposition:
= +
= 2 cos sin
Standing wave equation!
x (m)
68
**FOR EXTRA KNOWLEDGE ONLY.
Derivation of standing wave equation
y1 = Asin (ω t + k x) (to left )
y2 = Asin (ω t − k x) (to right )
y = y1 + y2
y = Asin (ω t + k x) + Asin (ω t − k x)
AB
From the Identity Trigonometry : sin( A ± B) = sin Acos B ± co
= (sin cos + cos sin ) + (s
= (Standing wave eq
Where; 2A cos kx – amplitude of standing wave at any x value
A – amplitude of individual progressive wave
2A – maximum amplitude of standing wave
os Asin B
sin cos − cos sin )
quation)
69
y
0
points of zero displacement – No
points of max displacement – an
Distance between 2 consecutive
Distance between consecutive n
λ is 2 times distance between co
x
ode (N)
ntinodes (AN)
e nodes or antinodes = λ/2
nodes and antinodes = λ/4
onsecutive nodes and antinodes.
70
• Node (N) is defined as a point at
the destructive interference occurr
• Antinode (A) is defined as a point
where the constructive interferenc
The pattern of the standing wave i
particles along the medium are diffe
appear at particular distance and dete
wave.
which the displacement is zero where
red.
at which the displacement is maximum
ce occurred.
is fixed hence the amplitude of each
erent. Thus, the nodes and antinodes
ermine by the equation of the standing
71
EXAMPLE 7.5.1 :
Transverse waves travel along a stretched string at speed 80 m
produced on the string. Determine the distance between :
(a) 2 consecutive nodes
(b) A node & an antinodes nearest to it.
Solution:
From: v = f λ
λ= v = 80 = 0.40 m
200
f
(a) Distance between 2 consecutive nodes = λ = 0.4 = 0.2
22
(b) Distance between a node & an antinode = λ = 0.4 =
44
m s–1 & frequency 200 Hz. standing waves are
*
*
2m
0.1 m
72
EXAMPLE 7.5.2 : Two identical waves are traveling towa
1 = 5 sin 2 + 2
5 3
(a) Write the equation of standing wave produced.
(b) Find the amplitude of a particle which is located at x = 2.6
(c) Find the position of nodes and antinodes.
(d) Calculate the speed of this wave.
Solution:
(a) From equation given we know that: = 2 and k = 2
5 3
Standing wave general equation:
=
(2x5)
= 10 cos 2 sin 2
3 5
ards each other.
2 = 5 sin 2 − 2
5 3
6 m.
(b) Amplitude of standing wave = 10 cos 2
3
= 10 cos 2 (2.6)
3
= 6.7 m
73
(c) Positions of nodes zero displacement
Positions of antiodes maximum displacement
Calculate λ by using = 2 where k= 2
λ 3
λ= 2 = 2 = 3
2
3
Let’s sketch y-x graph at t=0s: = 10 cos 2
3
( )
10
0 0.75 1.5 2.25 3.0 3.75 4.5 5.25 6.0
−10
= 0.75m, 2.25m, 3.75m, ⋯
= 0m, 1.5m, 3.0m, ⋯
(d) Speed of wave, =
: = 2
5
Since = 2
so = = 0.2
2
∴ =
= 0.2 (3)
= . −
(m)
⋯
⋯
74
EXAMPLE 7.5.3 :
Two harmonic waves are represented by the equations below
y1(x,t) = 3sin (πt +πx) y2(x,t) = 3sin (π
where y1, y2 and x are in centimetres and t in seconds.
a. Determine the amplitude of the new wave.
b. Write an expression for the new wave when both waves a
Solution:
(a) Amplitude of new wave: A=2a =2(3) = 6 cm
(b) By applying the principle of superposition, thus
y = y1(x,t)+ y2(x,t)
y = 3sin (πt +πx)+ 3sin (πt −πx)
y = 6cosπx sin πt
where y and x in centimetres and t in second