This chapter discusses sound waves and their properties. It defines sound as a longitudinal mechanical wave that travels through an elastic medium such as air. The chapter covers how the speed of sound depends on factors like the density and elasticity of the medium. Formulas are provided for calculating the speed of sound in solids, liquids, gases and air at different temperatures. It also examines the characteristic frequencies of open and closed pipes based on boundary conditions and the relationships between wavelength, frequency and velocity for sound waves.

1. Chapter 7 – Sound Waves
PHYS 243
Fall 2020-2021

2. Objectives: After completion of this
module, you should be able to:
• Define sound and solve problems
relating to its velocity in solids,
liquids, and gases.
• Use boundary conditions to apply
concepts relating to frequencies
in open and closed pipes.

3. Definition of Sound
Source of sound:
a tuning fork.
Sound is a longitudinal
mechanical wave that
travels through an
elastic medium.
Many things vibrate in
air, producing a sound
wave.

4. Is there sound in the forest
when a tree falls?
Based on our definition,
there IS sound in the
forest, whether a human
is there to hear it or not!
Sound is a physical
disturbance in an
elastic medium.
The elastic medium (air) is required!

5. Sound Requires a Medium
Evacuated Bell Jar
Batteries
Vacuum pump
The sound of a ringing bell diminishes as air leaves
the jar. No sound exists without air molecules.

6. Graphing a Sound Wave.
Sound as a pressure wave
The sinusoidal variation of pressure with distance is a
useful way to represent a sound wave graphically.
Note the wavelengths l defined by the figure.

7. Factors That Determine the Speed of Sound.
Longitudinal mechanical waves (sound) have a
wave speed dependent on elasticity factors and
density factors. Consider the following examples:
A denser medium has greater
inertia resulting in lower wave
speeds.
A medium that is more elastic
springs back quicker and
results in faster speeds.
steel
water

8. Speeds for different media
Y
v

= Metal rod
Young’s modulus, Y
Metal density, 
4
3
B S
v

+
= Extended
Solid
Bulk modulus, B
Shear modulus, S
Density, 
B
v

=
Fluid
Bulk modulus, B
Fluid density, 

9. Example 1: Find the speed of
sound in a steel rod.
vs = ?
11
3
2.07 x 10 Pa
7800 kg/m
Y
v

= = v =5150 m/s
 = 7800 kg/m3
Y = 2.07 x 1011 Pa

10. Speed of Sound in Air
For the speed of sound in air, we find that:
P RT
B P and
M


= =
 = 1.4 for air
R = 8.34 J/kg mol
M = 29 kg/mol
B P
v

 
= =
RT
v
M

=
Note: Sound velocity increases with temperature T.

11. Example 2: What is the speed of sound
in air when the temperature is 200C?
-3
(1.4)(8.314 J/mol K)(293 K)
29 x 10 kg/mol
RT
v
M

= =
Given:  = 1.4; R = 8.314 J/mol K; M = 29 g/mol
T = 200 + 2730 = 293 K M = 29 x 10-3 kg/mol
v = 343 m/s

12. Dependence on Temperature
RT
v
M

=
Note: v depends on
Absolute T:
Now v at 273 K is 331 m/s.
, R, M do not change, so
a simpler formula might be:
T
331 m/s
273K
v =
Alternatively, there is the approximation using 0C:
0
m/s
331 m/s 0.6
C
c
v t
 
= +  
 

13. Example 3: What is the
velocity of sound in air
on a day when the temp-
erature is equal to 270C?
Solution 1:
T
331 m/s
273K
v =
300 K
331 m/s
273K
v =
T = 270 + 2730 = 300 K;
v = 347 m/s
v = 347 m/s
Solution 2: v = 331 m/s + (0.6)(270C);

14. Musical Instruments
Sound waves in air
are produced by the
vibrations of a violin
string. Characteristic
frequencies are based
on the length, mass,
and tension of the
wire.

15. Vibrating Air Columns
Just as for a vibrating string, there are characteristic
wavelengths and frequencies for longitudinal sound
waves. Boundary conditions apply for pipes:
Open pipe
A A
The open end of a pipe must be
a displacement antinode A.
Closed pipe
A
N
The closed end of a pipe must
be a displacement node N.

16. Velocity and Wave Frequency.
The period T is the time to move a distance of one
wavelength. Therefore, the wave speed is:
1
but so
v T v f
T f
l
l
= = =
The frequency f is in s-1 or hertz (Hz).
The velocity of any wave is the product of the
frequency and the wavelength:
v f l
=
v
f
l
=

17. Possible Waves for Open Pipe
L 2
1
L
l =
Fundamental, n = 1
2
2
L
l =
1st Overtone, n = 2
2
3
L
l =
2nd Overtone, n = 3
2
4
L
l =
3rd Overtone, n = 4
All harmonics are
possible for open pipes:
2
1, 2, 3, 4 . . .
n
L
n
n
l = =

18. Characteristic Frequencies
for an Open Pipe.
L 1
2
v
f
L
=
Fundamental, n = 1
2
2
v
f
L
=
1st Overtone, n = 2
3
2
v
f
L
=
2nd Overtone, n = 3
4
2
v
f
L
=
3rd Overtone, n = 4
All harmonics are
possible for open pipes:
1, 2, 3, 4 . . .
2
n
nv
f n
L
= =

19. Possible Waves for Closed Pipe.
1
4
1
L
l =
Fundamental, n = 1
1
4
3
L
l =
1st Overtone, n = 3
1
4
5
L
l =
2nd Overtone, n = 5
1
4
7
L
l =
3rd Overtone, n = 7
Only the odd
harmonics are allowed:
4
1, 3, 5, 7 . . .
n
L
n
n
l = =
L

20. Possible Waves for Closed Pipe.
1
1
4
v
f
L
=
Fundamental, n = 1
3
3
4
v
f
L
=
1st Overtone, n = 3
5
5
4
v
f
L
=
2nd Overtone, n = 5
7
7
4
v
f
L
=
3rd Overtone, n = 7
Only the odd
harmonics are allowed:
L
1, 3, 5, 7 . . .
4
n
nv
f n
L
= =

21. Example 4. What length of closed pipe is needed
to resonate with a fundamental frequency of 256
Hz? What is the second overtone? Assume that
the velocity of sound is 340 M/s.
Closed pipe
A
N
L = ?
1, 3, 5, 7 . . .
4
n
nv
f n
L
= =
1
1
(1) 340 m/s
;
4 4 4(256 Hz)
v v
f L
L f
= = = L = 33.2 cm
The second overtone occurs when n = 5:
f5 = 5f1 = 5(256 Hz) 2nd Ovt. = 1280 Hz

22. Summary of Formulas For
Speed of Sound
Y
v

=
Solid rod
4
3
B S
v

+
=
Extended Solid
B
v

=
Liquid
RT
v
M

=
Sound for any gas:
0
m/s
331 m/s 0.6
C
c
v t
 
= +  
 
Approximation
Sound in Air:

23. Summary of Formulas (Cont.)
v
f
l
=
v f l
=
For any wave:
Characteristic frequencies for open and closed pipes:
1, 2, 3, 4 . . .
2
n
nv
f n
L
= = 1, 3, 5, 7 . . .
4
n
nv
f n
L
= =
OPEN PIPE CLOSED PIPE

24. CONCLUSION: Chapter 22
Sound Waves