about 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