πt −πx)
are superimposed.
ds.
75
7.5 (c) Compare between progressive wave
Progressive waves
produced by a source of produced
disturbance identical
direction.
wave profile moves away from the Wave pro
source of disturbance where the
energy is transmitted along energy is
direction of wave propagation. vibratory
Nodes are not found. All particles Nodes are
vibrate. nodes do
es and standing waves
Standing waves
d by superposition of 2
waves travel in opposite
.
ofile fixed in the region
ese waves meet
s retained within the
y motion of the wave
e found. Particles at
o not vibrate at all.
76
Exercise
1. The expression of a standing wave is given b
y = 0.3cos 0.5πx sin 60
where y and x in meters and t in seconds.
a. Write the expression for two progressiv
wave above.
b. Determine the wavelength, frequency,
progressive waves.
ANS. :
by
0πt
ve waves resulting the standing
amplitude and velocity for both
DIY, 4 m, 30 Hz, 0.15 m, 120 m s−1
77
7.6 : Application of standing
Learning Outcome:
a) Solve problems related to the fund
(i) stretched string, =
2
(ii) air columns , =
2
b) Use wave speed in a stretched string
g waves
damental and overtone frequencies for:
, =
4
g, =
78
7.6 (a) Solve problems related to the fundam
In general, standing waves ( standing waves ) are
as guitar when plucked, bowed or struck.
They are set up in the air in an organ pipe, a trump
mental and overtone frequencies
set up in the strings of musical instruments such
pet 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 str
then released.
• A wave that travels down a rope gets reflected at the
fixed, then the wave pulse is being inverted when r
• The reflections at the ends of the string cause two w
travel in opposite directions along the string.
• The incident waves & reflected waves from the fixe
retched between 2 fixed points is plucked and
e rope’s end. Since the end of the rope is
reflected.
waves of equal amplitude and wavelength to
ed ends interfere produce standing waves.
80
• Various type of
the tension in th
• Each modes ha
than the fundam
• These modes ar
• The fundamenta
standing wave t
(a)Fundamental mode or 1st Harmonic of a stretc
A
Fr
NN
= By
modes can be produced in the string – depends on
he string, how & where the string is plucked.
ave their corresponding frequency that are higher
mental frequency.
re called Overtones.
al frequency is simply the lowest frequency for a
to form and it also known as 1st Harmonic.
ched string
rom the figure, we know that : = 1 = 2
y using the speed of wave equation, = 1 1
1 =
= 1st Harmonic
81
(b) 1st Overtone or 2nd Harmonic : F
AA N By
N
N
We
From the figure, we know that :
=
y using the speed of wave equation, = 2 2
2 =
= 2nd Harmonic
e can also simplify the equation:
= 2 2
= 2 2nd Harmonic
82
(c) 2nd Overtone or 3rd Harmonic :
A AA
= NB
N
NN
W
From the figure, we know that :
= 3 3 =
2
By using the speed of wave equation, = 3 3
3 = =
= 3rd Harmonic
We can also simplify the equation:
=
= 3 3rd Harmonic
83
=
=
=
** the collection of all possible vibration modes is cal
series & n is called the harmonic number of the nth har
In General :
The frequencies of various
modes created in a string :
= =
=
= or
= OR = n = 1, 2, 3,
= 2
=
OR
= 3
lled the Harmonic
rmonic.
84
To calculate speed in string: Wave speed in string,
since =
fn = n
2L
Equation shows, the fundam
wave in string as well as the
1. length of the string, L
2. tension , T
3. linear mass density of t
= where L = length of the string
n = number of harmonic
T = tension of the string
μ = mass per unit length of string
( T )
L
µ
mental frequency of standing
other harmonics depends on:
the string, μ.
85
EXAMPLE 7.6.1 :
The tension in a stretched string of length 50 cm, mas
(a)The speed of the transverse waves travelling along
(b) The fundamental frequency
(c) The wavelength of the progressive wave which mo
Solution: Given : L = 50 cm = 0.5 m; m = 1.0 g = 1×10
(a) The mass per unit length μ of the string is µ=m = 1×
L
The speed of the transverse wave in string :
v= T
µ
= 100
2 ×10 −3
= 223.61m s−1
ss 1.0 g is 100 N. When the string vibrates, determine
g the string.
ove along the wire when at 2nd overtone ?
0–3 kg; T = 100 N
×10−3 = 2×10−3 kg m−1
0.5
86
(b) Fundamental frequency :
f1= v
2L
= 223.6
2 (0.5
= 223.6
(c) 2nd Overtone Ξ 3rd Harmonic, thus n =
from : fn = n f 1
f 3 = (3) (223 .61)
f 3 = 670.83
61 from: v = f λ
5)
61 Hz λ= v 223.61
= 670.83
=3 f
1
)
= .
87
ii) Air Column (Closed Pipe)
• If a vibrating tuning fork is placed at the open end of a
• The sound waves travel in the air column towards the
• They are reflected at the closed end. Reflected wave
standing waves.
• The fundamental mode formed in closed pipe : N
• At the closed end of the pipe, air molecules can’t vibr
• At the open end air molecules free to vibrate with m
a long hallow pipe as shown below.
e closed end.
es interfere with the incident waves produced
A
L
rate : Nodes (N)
maximum displacement : Antinode (A)
88
(a)Fundamental mode or 1st Harmonic
Let : L – length of the closed pipe
v – speed of sound wave
=
4
NA
From the figure, we know that :
= ⇒ =
4
By using the speed of wave equation,
= 1
1 =
= 1st Harmonic
89
(b)1st Overtone or 3rd Harmonic
Let : L – length of the closed pipe
v – speed of sound wave
= 3
4
N AN A
From the figure, we know that :
= 3 ⇒ =
4
By using the speed of wave equation,
=
= 4
=
3
= 3rd Harmonic
We can also simplify the equation:
=
= 3 3rd Harmonic
90
(c) 2nd Overtone or 5th Harmonic
Let : L – length of the closed pipe
v – speed of sound wave
= 5
4
N A N AN A
From the figure, we know that :
= 5 ⇒ =
4
By using the speed of wave equation,
= 2
4
= = 5
= 5th Harmonic
We can also simplify the equation:
=
= 5 5th Harmonic
91
Fundamental =
(1st Harmonic)
1st Overtone =
(3rd Harmonic)
2nd Overtone =
(5th Harmonic)
!!!For closed pipe, only odd harmonics exis
In General :
The frequencies of various
modes created in a string :
=
=
= or
=
3 OR
where n = odd numbers
= 3 n = 1, 3, 5, 7, 9, …
v = speed of sound wave
= L = length of closed pipe
f1 = fundamental frequency
OR
= 5
st.
92
EXAMPLE 7.6.2 :
The length of a closed pipe is 15.0 cm. If the speed of so
frequencies for the sound emitted when one blows acro
Solution: 1st Ov
3 lowest frequencies that can be produced :
2nd O
For the Fundamental note ( 1st Harmonic ) : ∴ The
f 1 = v = 330 = 550 Hz
4L 4(0.15)
ound in air is 330 m s–1, find the 3 lowest
oss the opened end of the pipe.
vertone ( or 3rd Harmonic ) : ( n = 3 )
from : fn = n f 1
3 = 3 (550)
=1650 Hz
Overtone ( or 5th Harmonic ) : ( n = 5 )
from : fn = n f 1
5 = 5 (550)
3 lowest frequencies are 550 Hz , 1650 Hz , 2750 Hz
93
iiI) Air Column (Open pipe Pipe)
• A standing wave is produced in an open pipe
the pipe.
• Molecules of air at both the opened ends free
• At midpoint amplitude is zero – node ( N )
when you blow across one end of
to vibrate with maximum amplitude – antinodes ( A )
94
(a)Fundamental mode or 1st Harmonic
Let : L – length of the closed pipe
v – speed of sound wave
=
2
A NA
From the figure, we know that :
= ⇒ =
2
By using the speed of wave equation,
= 1
1 =
= 1st Harmonic
95
(b)1st Overtone or 2nd Harmonic A
Let : L – length of the closed pipe
v – speed of sound wave
=
A N AN
From the figure, we know that :
=
By using the speed of wave equation,
=
=
=
= 2nd Harmonic
We can also simplify the equation:
=
= 2 2nd Harmonic
96
(c) 2nd Overtone or 3rd Harmonic
Let : L – length of the closed pipe
v – speed of sound wave
= 3
2
AN A N A N A
From the figure, we know that :
= 3 ⇒ =
2
By using the speed of wave equation,
=
= 2
=
3
= 3rd Harmonic
We can also simplify the equation:
=
= 3 3rd Harmonic
97
Fundamental =
(1st Harmonic) =
=
1st Overtone
(2nd Harmonic)
2nd Overtone
(3rd Harmonic)
In General :
The frequencies of various
modes created in a string :
=
=
= or
= =
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 bo
is 594 Hz.
(a) How long is this pipe? (b)(
(b) If one end is now closed, find
i. the wavelength and
ii. frequency of the new fundamental.
Solution: 1 =
2
(a) For an open pipe,
Given that 1 =594 Hz (b)
L= v
2 f1
= 344 ms-1
2(594 Hz)
= 0.290 m
oth ends
(i) There is a node at one end, an antinode at the
other end and no other nodes or antinodes in
between, so
= 1
4
λ=1 4=L 4(0.290) = 1.16 m
)(ii) f1 = v
4L
1 From question(a) 2 =594 Hz
= 2 2
= 1 (594Hz) 99
2
= 297 Hz
EXAMPLE 7.6.3 :
Find the fundamental frequency and the frequency of t
(a) if the pipe is open at both ends and
(b) if the pipe is closed at one end.
Use v = 344 m/s.
Solution:
A pipe closed at one end is a closed pipe.
For the open pipe n =1, 2, and 3 for the first three harmon
For the closed pipe n =1, 3, and 5. (odd number only)
(a) Fundamental frequency of open pipe: f1 = v
2L
f1 = 344ms-1
2(0.450m)
= 382Hz