CHAPTER 7: OSCILLATIONS & WAVES

7.4 (d) Distinguish between particle vibrational velocity and wave propagation velocity.



Undisturbed
position of string



• A transverse wave on a string is moving to the right with a 50
constant speed, vwave.

• A string particle moves up and down in SHM about the
undisturbed position of the string.

• The speed of the particle vy changes from time to time
as the wave passes.

Particle vibrational velocity Wave propagation velocity

Velocity of particle ( ) which oscillate about The wave velocity ( ) is the velocity at which wave crests
equilibrium position in simple harmonic motion. (or other part of waveform) move forward. It can be defined
as the distance travelled by a wave profile per unit time.

By differentiating the displacement equation of the Figure below shows a progressive wave profile moving to the

wave, thus ; right. v

vy = dy = d [ A sin (ω t ± k x)]
dt dt
λ

vy = Aω cos (ω t ± k x) The trough travels a distance of one wavelength, in a time
equal to one period, T. Thus, the wave velocity is:

v = distance v = λ where T = 1
time T f

∴ =

Particle vibrational velocity, vy is not constant and varies The wave velocity , vwave is constant
with time, t .

51

7.4 (e) Use particle vibrational velocity, vy = Aω cos (ω t ± k x)

(f) Use wave propagation velocity, =

EXAMPLE 7.4.4 :

A progressive wave equation is given as : y = 3sin (8π t + 0.4 x) cm

Find : (b) from : k = 2π
(a) Particle velocity, vy λ
(b) Velocity of the wave v
λ = 2π = 2π = 5π
Solution:
k 0.4
(a) vy = dy = Aω cos (ω t + k x)
from :ω = 2π f
dt
= (3) (8π ) cos (8π t + 0.4 x) f = ω = 8π = 4 Hz
2π 2π
= 24π cos (8π t + 0.4 x) cm s−1
∴ wave velocity, v = f λ
= 4 (5π ) 52

= 20π cm s−1

7.4 (g) Analyse the graphs of: (i) displacement–time, y-t
(ii) displacement–distance, y-x

i) Graph of displacement against time, y-t

The graph shows the displacement of any one particle in the wave at any particular distance, x from the origin.

For example, consider the equation of the wave is :

y = Asin (ωt − kx)

When x = 0 , thus

y = Asin (ωt − k(0))
y = Asin (ωt)

Hence, the displacement-time, y-t graph A
of = ( ) is :
0

−A
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 wave at any particular time, t.
For example, consider the equation of the wave is :

y = Asin (ωt − kx)

When t = 0 , thus y

y = Asin (ω(0)− kx)
y = Asin (− kx)

y = −Asin (kx)

Hence, the displacement-distance, y-x graph A v
of = − ( ) is :

0 λ λ 3λ x
2λ
−A 2 2
54

EXAMPLE 7.4.5 :

y (cm) 1.0 x (cm)

3 2.0

0

−3

Figure above shows a displacement, y against distance, x graph after time, t for the
progressive wave which propagates to the right with a speed of 50 cm s−1.

a. Determine the wave number and frequency of the wave.
b. Write the expression of displacement as a 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 − 200πx)
y(x,t ) = 0.03sin (100πt − 200πx)

Where y and x in metres and t in seconds

56

EXAMPLE 7.4.6 :

A progressive wave is represented by the equation , = − where y and x are in centimeters and t in
seconds.

a. Determine the angular frequency, the wavelength, the period, the frequency and the wave speed.

b. Sketch the displacement against distance graph for progressive wave above in a range of 0 ≤ x ≤ λ at time, t = 0 s.

c. Sketch the displacement against time graph for the particle at x = 0 in a range of 0 ≤ t ≤ T.

d. Sketch the displacement against time graph for the particle at x = 0.5λ in a range of 0 ≤ t ≤ T.

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, (ii) k = π cm−1 and k = 2π 2π = π
(a) By comparing the general equation λ λ
(i) ω = π rad s−1
Therefore, = 2 cm

, = − (iii) The period : (iv) The frequency and wave speed:

with = 2 = 1 = 1 f = 0.5 Hz
2
, = − By applying the equation of wave speed v = λf
2
= = 2 × 10−2 0.5

T =2s = 1 × 10−2 m s−1 57

(b) At time, t = 0 s, the equation of displacement as a function of distance, x is given by

, 0 = 2 sin 0 −

, 0 = 2 sin − y (cm)
, 0 = −2 sin
2
, 0 = −2 sin

**Note:

Calculate λ first: 2
λ
since = and = 0 x (cm)
∴λ= 2 2 2
= = 12

−2

58

(c) The particle at distance, x = 0 , the equation of displacement as a function of time, t is given by

0, = 2 sin − 0
0, = 2 sin

Hence the displacement, y against time, t graph is

**Note:

y (cm) Calculate first: 2

2 y(0,t ) = 2sin (πt ) since = and =
2
∴ = 2 = 2 =


t (s)

0 12

−2

59

(d) The particle at distance, x = 0.5λ and λ = 2 cm thus x = 1 cm. Therefore, the equation of displacement
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 range of 0 ≤ t ≤ T is

t 0 0.5 1 1.5 0 y (cm)
y 0 -2 0 2 0
2

0 t (s)

12

−2

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. Determine
a) the amplitude, wavelength and frequency of the
wave.
b) the velocity with which the wave moves along the
string.
c) the displacement of a particle located at x = 22.5 cm
and t = 18.9 s.

ANS: 0.327 cm, 8.71 cm, 0.433 Hz; 0.0377 m s−1; −0.192 cm

62

7.5 : Superposition of Waves

Learning Outcome:

a) State the principle of superposition of waves for the constructive and
destructive interferences.

b) Use the standing wave equation = 2 cos sin
c) Compare between progressive waves and standing waves.

63

7.5 (a) State the principle of superposition of waves for the constructive and destructive
interferences.

Superposition of waves means when two waves meet in the same region, they interfere.

Principle of superposition of waves: At any time, the combined waveform of two or more
interfering waves is given by the sum of the displacements
of the individual waves at each point in the medium.

 
y = y1 + y2

Where y1 and y2 are the displacement of individual pulses at that point.

64

Interference is defined as the interaction (superposition) of two or more wave motions.

1. Constructive interference : wave pulse 1 wave pulse 2

the vertical displacements of the Before Wave pulse 1 and Wave pulse 2 are
two pulses are in the same interference
direction, and the amplitude of A A travelling against each other. Both
the combined waveform is greater have the same amplitude, A.
than that of either pulse.
During A+A The amplitude of the wave doubles
• It occurs wave pulse 1 and 2 have the interference (A + A) when both waves interfered.
same wavelength, frequency and in 2A
phase each other. Resultant displacement, y = 2A

After A Both waves travel with the initial
interference A amplitude, A.

65

2. Destructive interference : wave pulse 1

one pulse has a negative Before A Wave pulse 1has positive amplitude while
displacement, the two pulses tend interference wave pulse 2 has negative amplitude.
to cancel each other when they
overlap, and the amplitude of the -A
combined waveform is smaller
than that of either pulse. wave pulse 2

• It occurs wave pulse 1 and 2 have the During A+(-A) the sum of the amplitude for wave A+(-A) is zero .
same wavelength, frequency and interference
antiphase each other. Resultant displacement, y = 0

After Both waves propagate with initial
interference A amplitude.

--A

66

7.5 (b) Use the standing wave equation =

• wave whose profile do not advance & no energy is transferred from one end of the
medium to the other end.

• Standing wave appear to be vibrating vertically without traveling horizontally.
• produced by the superposition of two progressive waves of equal amplitude and

frequency, traveling in opposite direction.
• also known as stationary waves.

67

By considering the wave functions for two progressive waves: y1 = Asin(ωt − kx) and y2 = Asin(ωt + kx)

y (m) Applying principle of superposition:
+ 2A
= +
standing wave: = 2 cos sin
Standing wave equation!
− 2A
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 ± cos Asin B
= (sin cos + cos sin ) + (sin cos − cos sin )

= (Standing wave equation)

Where; 2A cos kx – amplitude of standing wave at any x value
A – amplitude of individual progressive wave

2A – maximum amplitude of standing wave

69

y
0x

 points of zero displacement – Node (N)
 points of max displacement – antinodes (AN)
 Distance between 2 consecutive nodes or antinodes = λ/2
 Distance between consecutive nodes and antinodes = λ/4
 λ is 2 times distance between consecutive nodes and antinodes.

70

• Node (N) is defined as a point at which the displacement is zero where
the destructive interference occurred.

• Antinode (A) is defined as a point at which the displacement is maximum
where the constructive interference occurred.

The pattern of the standing wave is fixed hence the amplitude of each
particles along the medium are different. Thus, the nodes and antinodes
appear at particular distance and determine by the equation of the standing
wave.

71

EXAMPLE 7.5.1 : *

Transverse waves travel along a stretched string at speed 80 m s–1 & frequency 200 Hz. standing waves are
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 m

22

(b) Distance between a node & an antinode =λ = 0.4 = 0.1 m
4
4

72

EXAMPLE 7.5.2 : Two identical waves are traveling towards each other.

1 = 5 sin 2 + 2 2 = 5 sin 2 − 2
5 3 5 3

(a) Write the equation of standing wave produced.
(b) Find the amplitude of a particle which is located at x = 2.6 m.
(c) Find the position of nodes and antinodes.
(d) Calculate the speed of this wave.

Solution: (b) Amplitude of standing wave = 10 cos 2
3
2 2
(a) From equation given we know that: = 5 and k = 3 = 10 cos 2 (2.6)
3
Standing wave general equation:
= 6.7 m

=

(2x5)

= 10 cos 2 sin 2
3 5
73

(c) Positions of nodes zero displacement (d) Speed of wave, =

Positions of antiodes maximum displacement 2
5
Calculate λ by using = 2 where k= 2 : =
λ 3
Since = 2
2 2 = 3
λ= =
2 2
3 so = = 0.2

Let’s sketch y-x graph at t=0s: = 10 cos 2
3
∴ =
( ) = 0.2 (3)

10 = . −

0 0.75 1.5 2.25 3.0 3.75 4.5 5.25 6.0 (m)

−10

= 0.75m, 2.25m, 3.75m, ⋯
= 0m, 1.5m, 3.0m, ⋯

74

EXAMPLE 7.5.3 :

Two harmonic waves are represented by the equations below

y1(x,t) = 3sin (πt +πx) y2(x,t) = 3sin (πt −πx)

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 are superimposed.

Solution:

(a) Amplitude of new wave: A=2a =2(3) = 6 cm

(b) By applying the principle of superposition, thus

y = y1(x,t)+ y2(x,t)

y = 3sin (πt +πx)+ 3sin (πt −πx)

y = 6cosπx sin πt

where y and x in centimetres and t in seconds.

75

7.5 (c) Compare between progressive waves and standing waves

Progressive waves Standing waves

produced by a source of produced by superposition of 2
disturbance identical waves travel in opposite
direction.
wave profile moves away from the Wave profile fixed in the region
source of disturbance where these waves meet
energy is transmitted along energy is retained within the
direction of wave propagation. vibratory motion of the wave
Nodes are not found. All particles Nodes are found. Particles at
vibrate. nodes do not vibrate at all.

76

Exercise

1. The expression of a standing wave is given by

y = 0.3cos 0.5πx sin 60πt

where y and x in meters and t in seconds.
a. Write the expression for two progressive waves resulting the standing
wave above.
b. Determine the wavelength, frequency, amplitude and velocity for both
progressive waves.

ANS. : DIY, 4 m, 30 Hz, 0.15 m, 120 m s−1

77

7.6 : Application of standing waves

Learning Outcome:

a) Solve problems related to the fundamental and overtone frequencies for:

(i) stretched string, =
2

(ii) air columns , = , =
2 4

b) Use wave speed in a stretched string, =


78

7.6 (a) Solve problems related to the fundamental and overtone frequencies
In general, standing waves ( standing waves ) are set up in the strings of musical instruments such
as guitar when plucked, bowed or struck.

They are set up in the air in an organ pipe, a trumpet or a clarinet when air is blown over the top.

79

i) Standing wave in
stretched string

• Standing wave can be produced when a string stretched between 2 fixed points is plucked and
then released.

• A wave that travels down a rope gets reflected at the rope’s end. Since the end of the rope is
fixed, then the wave pulse is being inverted when reflected.

• The reflections at the ends of the string cause two waves of equal amplitude and wavelength to
travel in opposite directions along the string.

• The incident waves & reflected waves from the fixed ends interfere produce standing waves.

80

• Various type of modes can be produced in the string – depends on
the tension in the string, how & where the string is plucked.

• Each modes have their corresponding frequency that are higher
than the fundamental frequency.

• These modes are called Overtones.

• The fundamental frequency is simply the lowest frequency for a
standing wave to form and it also known as 1st Harmonic.

(a)Fundamental mode or 1st Harmonic of a stretched string

A From the figure, we know that : =  1 = 2

N N

= By using the speed of wave equation, = 1 1


1 =

=  1st Harmonic

81

(b) 1st Overtone or 2nd Harmonic : From the figure, we know that :
AA
=
N
N N By using the speed of wave equation, = 2 2

2 =

=  2nd Harmonic


We can also simplify the equation:

= 2 2

= 2  2nd Harmonic

82

(c) 2nd Overtone or 3rd Harmonic : From the figure, we know that :
A AA
= 3 3  =
2

= By using the speed of wave equation, = 3 3

N N

NN 3 = =


=  3rd Harmonic


We can also simplify the equation:

=


= 3  3rd Harmonic

83

In General :

The frequencies of various
modes created in a string :

= =

=

= or

= = n = 1, 2, 3, 
OR

= 2

=

=
OR

= 3

** the collection of all possible vibration modes is called the Harmonic
series & n is called the harmonic number of the nth harmonic.

84

To calculate speed in string: Wave speed in string, = where L = length of the string
n = number of harmonic
T = tension of the string
μ = mass per unit length of string

since =


fn = n ( T )
2L
µ

Equation shows, the fundamental frequency of standing
wave in string as well as the other harmonics depends on:

1. length of the string, L
2. tension , T
3. linear mass density of the string, μ.

85

EXAMPLE 7.6.1 :

The tension in a stretched string of length 50 cm, mass 1.0 g is 100 N. When the string vibrates, determine
(a)The speed of the transverse waves travelling along the string.
(b) The fundamental frequency
(c) The wavelength of the progressive wave which move along the wire when at 2nd overtone ?

Solution: Given : L = 50 cm = 0.5 m; m = 1.0 g = 1×10–3 kg; T = 100 N

(a) The mass per unit length μ of the string is µ=m = 1×10−3 = 2×10−3 kg m−1
L 0.5

The speed of the transverse wave in string :

v= T
µ

= 100
2 ×10 −3

= 223.61m s−1 86

(b) Fundamental frequency :

f1= v
2L

= 223.61
2 (0.5)

= 223.61 Hz

(c) 2nd Overtone Ξ 3rd Harmonic, thus n = 3 from: v = f λ

from : fn = n f 1 λ= v 223.61
= 670.83
f 3 = (3) (223 .61) f
f 3 = 670.83
= .

87

ii) Air Column (Closed Pipe)
• If a vibrating tuning fork is placed at the open end of a long hallow pipe as shown below.

• The sound waves travel in the air column towards the closed end.

• They are reflected at the closed end. Reflected waves interfere with the incident waves produced
standing waves.

• The fundamental mode formed in closed pipe : N A

L

• At the closed end of the pipe, air molecules can’t vibrate : Nodes (N)
• At the open end air molecules free to vibrate with maximum displacement : Antinode (A)

88

(a)Fundamental mode or 1st Harmonic

Let : L – length of the closed pipe
v – speed of sound wave

By using the speed of wave equation,
4
=

= 1

NA 1 =


From the figure, we know that : =  1st Harmonic


= ⇒ =
4

89

(b)1st Overtone or 3rd Harmonic

Let : L – length of the closed pipe By using the speed of wave equation,
v – speed of sound wave

3 =
= 4 = 4

=

N AN A 3

=  3rd Harmonic


We can also simplify the equation:

=


From the figure, we know that : = 3  3rd Harmonic

= 3 ⇒ =
4
90

(c) 2nd Overtone or 5th Harmonic

Let : L – length of the closed pipe By using the speed of wave equation,
v – speed of sound wave

5 = 2
= 4 4
5
= =

N A N AN A

=  5th Harmonic


We can also simplify the equation:

=


From the figure, we know that : = 5  5th Harmonic

= 5 ⇒ =
4
91

In General :

The frequencies of various
modes created in a string :

Fundamental = =
(1st Harmonic)
=

= or

=
1st Overtone = 3 OR
(3rd Harmonic)
= 3 where n = odd numbers
2nd Overtone n = 1, 3, 5, 7, 9, …
(5th Harmonic) = v = speed of sound wave
L = length of closed pipe
= f1 = fundamental frequency
OR

= 5

!!!For closed pipe, only odd harmonics exist.

92

EXAMPLE 7.6.2 :

The length of a closed pipe is 15.0 cm. If the speed of sound in air is 330 m s–1, find the 3 lowest
frequencies for the sound emitted when one blows across the opened end of the pipe.

Solution: 1st Overtone ( or 3rd Harmonic ) : ( n = 3 )
3 lowest frequencies that can be produced :
from : fn = n f 1

3 = 3 (550)

=1650 Hz

2nd Overtone ( or 5th Harmonic ) : ( n = 5 )

from : fn = n f 1

5 = 5 (550)

For the Fundamental note ( 1st Harmonic ) : ∴ The 3 lowest frequencies are 550 Hz , 1650 Hz , 2750 Hz

f 1 = v = 330 = 550 Hz 93
4L 4(0.15)

iiI) Air Column (Open pipe Pipe)

• A standing wave is produced in an open pipe when you blow across one end of
the pipe.

• Molecules of air at both the opened ends free to vibrate with maximum amplitude – antinodes ( A )

• At midpoint amplitude is zero – node ( N )

94

(a)Fundamental mode or 1st Harmonic

Let : L – length of the closed pipe
v – speed of sound wave

= By using the speed of wave equation,
2

= 1

A NA 1 =


From the figure, we know that : =  1st Harmonic


= ⇒ =
2

95

(b)1st Overtone or 2nd Harmonic By using the speed of wave equation,

Let : L – length of the closed pipe =
v – speed of sound wave =
=
=
A
A N AN

=  2nd Harmonic


We can also simplify the equation:

=


From the figure, we know that : = 2  2nd Harmonic

=

96

(c) 2nd Overtone or 3rd Harmonic

Let : L – length of the closed pipe By using the speed of wave equation,
v – speed of sound wave

= 3 =
2 = 2
=

AN A N A N A 3

=  3rd Harmonic


We can also simplify the equation:

=


From the figure, we know that : = 3  3rd Harmonic

= 3 ⇒ =
2
97

In General :

The frequencies of various
modes created in a string :

Fundamental = =
(1st Harmonic)
=
1st Overtone
(2nd Harmonic) = or

2nd Overtone = =
(3rd Harmonic) OR

= 2 where n = integer numbers
n = 1, 2, 3, 4, …
= v = speed of sound wave
L = length of open pipe
= f1 = fundamental frequency
OR

= 3

98

EXAMPLE 7.6.2 :

The fundamental frequency of a pipe that is open at both ends

is 594 Hz. (b)(i) There is a node at one end, an antinode at the
(a) How long is this pipe? other end and no other nodes or antinodes in
(b) If one end is now closed, find between, so

i. the wavelength and

ii. frequency of the new fundamental.

Solution: 1 = 1
2 4
(a) For an open pipe, =

Given that 1 =594 Hz λ=1 4=L 4(0.290) = 1.16 m

L= v (b)(ii) f1 = v
2 f1 4L

= 344 ms-1 1 From question(a) 2 =594 Hz
2(594 Hz) = 2 2

= 0.290 m = 1 (594Hz)

2 99

= 297 Hz