1) A wave is a disturbance that transfers energy through a medium without permanent transfer of matter. This document discusses mechanical and electromagnetic waves, their characteristics such as wavelength and frequency, and how waves behave in different media like strings and pipes. 2) Resonance occurs when an object vibrates at its natural frequency, causing large oscillations. Standing waves can form in pipes and strings, with nodes and antinodes depending on boundary conditions and wavelength. 3) The Doppler effect describes the change in frequency heard by an observer due to relative motion between the source and observer. The pitch is higher if approaching and lower if receding due to the change in wavelength reaching the observer each second.

1. Chapter 16
Waves and Sound
Younes Sina

2. A wave is the motion of a disturbance in a medium.
The medium for ocean waves is water.
When a string, fixed at both ends, is given a vertical hit by a stick, a dent appears
in it that travels along the string. When it reaches an end point, it reflects and
inverts and travels toward the other end. The following figure shows the motion
of a single disturbance.

3. Types of Waves:
classification of waves :
Mechanical
Electromagnetic
Mechanical waves require matter for their transmission. Sound waves, ocean
waves, and waves on a guitar string are examples. Air, water, and metal
string are their media (matter), respectively.
Electromagnetic waves can travel both in vacuum and matter. If light (a wave
itself) could not travel in vacuum, we would not see the Sun. Light is an
electromagnetic wave. Radio waves, ultraviolet waves, and infrared waves
are all electromagnetic waves and travel in vacuum.

4. Waves are also classified as:
 Transverse
 Longitudinal
For a transverse wave the disturbance direction is perpendicular to the propagation
direction.
Water waves are transverse. Waves on guitar strings are also transverse.
For a longitudinal wave the disturbance direction is parallel to the propagation
direction. Waves on a slinky as well as sound waves are longitudinal.

6. Frequency ( f ):
Frequency ( f )
The number of full waveforms generated per second.
The SI unit for frequency is (1/s), or (s -1), called "Hertz (Hz)“.
Period ( T ):
Period is the number of seconds per waveform, or the number of seconds
per oscillation.
T = 1 / f
Relation between frequency ( f ) and the angular speed ( ω ):
ω = 2π f
ω is the number of radians per second
f is the number of turns per second
Each turn is 2π radians

7. Wavelength ( λ ) is the distance between two successive points on a wave that are
in the same state of oscillation.
Points A and B in the above Figure are the nearest or successive points that are both
the same amount passed the maximum and therefore in the same state of oscillation.

8. At the following Figure the distance between any node and the anti-node
next to it is λ/4.

9. Wave Speed ( v )
The wave speed is the distance a wave travels per second.
v = f λ
Example :
The speed of sound waves at STP conditions is 331 m/s.
Calculate the wavelength of a sound wave with a frequency of 1324 Hz at STP.
Solution:
v = f λ → λ = v/f
λ = (331m/s) / (1324/s) = 0.250 m

10. The Vibrating String
Nodes are points of zero oscillation and antinodes are points of maximum
oscillation as shown in following Figure.

11. Example:
In a 60.0-cm long violin string, three antinodes are
observed. Find the wavelength of the waves on it.
Solution: Each loop has a length of (60.0cm/3)=20.0 cm.
Each wavelength (a full sine wave) contains two of such
loops; therefore,
λ = 40.0 cm

12. Speed of Waves in a Stretched String:
The speed of waves in a stretched string depends on the tension F in the string as
well as the mass per unit length, μ, of the string as explained below:
The more a string is stretched, the faster waves travel in it.
The formula that relates tension F in the string and the waves speed, v, is:
mass per unit length

13. Example : A 120-cm guitar string is under a tension of 400 N. The mass of the
string is 0.480 grams. Calculate
(a) the mass per unit length of the string and
(b) the speed of waves in it
(c) In a diagram show the number of (1/2) λ that appear in this string if it is
oscillating at a frequency of 2083 Hz.
Solution:
(a) μ = M / L
μ = (0.480x10-3 kg) /1.20 m = 4.00x10-4 kg/m
(b) v = (F/μ)1/2
v = (400N / 4.00x10-4 kg/m)1/2 = 1000 m/s
(c) v = f λ → λ = v / f
λ = (1000 m/s) / (2083/s) = 0.480 m
(1/2)λ = 0.480 m/ 2 = 0.240 m
The number of (λ/2)'s that fit in the string length of 120 cm is 1.20 m/ 0.240 m
= 5.00

14. Resonance
In physics, resonance is the tendency of a system to oscillate with greater
amplitude at some frequencies than at others. Frequencies at which the response
amplitude is a relative maximum are known as the resonance frequencies.
"The phenomenon of making a body vibrate with its natural frequency under
the influence of another vibrating body having the same frequency is called
RESONANCE“
Two oscillatory (periodic) motions have the same period (or frequency) and
are also in phase.

15. The Resonance of Sound Waves in Pipes:
From music point of view, a pipe open at both ends is called an "open pipe".
A pipe, closed at one end only, is called a "closed pipe.“
Sound waves are longitudinal (their disturbance direction is parallel to their
propagation direction). In pipes, sound waves oscillate parallel to the pipe's
length.

16. 1) Closed Pipes:
At the closed end of a closed pipe, only a node can form, because the air
molecules (inside the pipe) that transmit sound waves have to bounce off the
closed end after collision.
They have to come to stop before bouncing off. Coming to stop means zero
state of oscillation at that closed end that results in node formation.
At the open end of a closed pipe; however, both nodes and antinodes are possible
to form depending on the wavelength of the sound waves as well as the pipe's length.

18. For a closed pipe, resonance occurs when the pipe's length is an odd multiple of ( λ/4 )
L1 = 1 λ/4
L3 = 3 λ/4
L5 = 5 λ/4
L7 = 7 λ/4

19. 2) Open Pipes:
An open pipe is open at both ends; therefore, both ends must form antinodes for
the pipe to be "in resonance."
The pipe length must be an even multiple of ( λ/4 ) for resonance.
For an open pipe, resonance occurs when pipe's length is an even
multiple of (λ/4 ).
L1 = 2 λ/4
L3 = 4 λ/4
L5 = 6 λ/4
L7 = 8 λ/4

20. The Speed of Sound in a Gas:
The speed of sound in a medium is a function of the physical properties of that
medium. The speed of sound in a gas is a function of:
gas temperature
Pressure
density
Example :
A tuning fork oscillating at a rate of 686 Hz is brought close to the open end of a
closed tube in a room at a certain temperature. The tube's length is changed
from 0.0 to 40.0 cm and two resonances (load sounds) are heard:
the first one, at a tube length of 12.5 cm, and next, at a tube length of 37.5 cm.
Calculate the speed of sound at that temperature.
v(T)=[331+0.6T] m/s T (in oC)

21. Solution:
The smallest length of a closed pipe for resonance is λ/4. The first resonance
length of 12.5 cm means λ/4 = 12.5 cm.
The second resonance occurring at a length of 37.5 cm means 3λ/4 = 37.5 cm
λ/4 = 12.5 cm
λ = 4(12.5cm) = 50.0 cm
3λ/4 = 37.5cm
λ = (4/3)(37.5 cm)
λ = 50.0 cm
Getting the same results should not be surprising because they are both
measurements of the same thing.
Knowing λ and the frequency of the waves: f = 686 Hz, the sound speed at that
temperature is
v = f λ
vsound = [686 (1/s)] (0.500m) = 343m/s
Speed of sound at STP conditions (0oC and 1atm pressure) is 331 m/s

22. The Doppler Effect:
When an ambulance is approaching us, a higher pitch sound is heard than
when it is going away from us. The reason is that when the sound source is
moving toward us, more number of wavelengths per second arrive at our
ears than when it is at rest. When the sound source (ambulance) is going
away from us, less number of wavelengths arrive at our ears than when the
source (ambulance) is at rest.

23. Cases (1) and (2) are when the observer is at rest and
the source is approaching or receding.
❷❶
Five cases for the frequency heard by the observer:

24. Cases (3) and (4) are when the source is at rest and the
observer is approaching or going away.
❹
❸

25. ❺
Case (5) is when both observer and source are moving,
either approaching or receding.

26. fo = the frequency heard by the observer
fs is the frequency of the sound source
vs is the speed of the source
vo is the speed of the observer
v is the speed of sound in the medium.
1) Denom. < Numerator →fo > fs
2) Denom. > Numerator →fo < fs
3) Denom. < Numerator →fo > fs
4) Denom. > Numerator →fo < fs
5) Choose (+) and (-) signs that
make sense

27. Case 5: If both the observer and the source are approaching, the highest
possible frequency is heard. To make the fraction the greatest, chose the (+) in
the numerator and the (-) in the denominator.
If both the observer and the source are receding, the lowest possible frequency
is heard. To make the fraction the least, chose the (-) in the numerator and the
(+) in the denominator.
If the source is chasing the observer, choose (-) in the numerator and (-) in the
denominator.
If the observer is chasing the source, choose (+) in the numerator and (+) in the
denominator.
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28. Example :
If an ambulance with its siren on at a frequency of 1350 Hz is approaching you
at a speed of 33.1 m/s at STP conditions, calculate
(a) the frequency you hear.
(b) If you are driving at a speed of 16.55 m/s toward the coming ambulance,
what frequency do you hear?
(c) If you are driving at a speed of 16.55 m/s away from the moving
ambulance, what frequency do you hear?
(d) What frequency do you hear when the ambulance passes your car and
continues in front of you?
(e) What frequency do you hear if both of you and the ambulance stop?
a
d
c
b
e

29. Solution:
(a) Case 1: fo = fs [ (Vo) / (V - Vs) ]
fo = 1350Hz [(331 + 0) /( 331 - 33.1 )] = 1500 Hz
(b) Case 5, source & observer moving toward each other: fo= fs[(V + Vo)/(V-Vs)]=
1575 Hz
(c) Case 5, source chasing the observer: fo = fs [(V-Vo)/(V-Vs)] = 1425 Hz
(d) Case 5, observer chasing the source: fo = fs [(V + Vo)/(V + Vs)] = 1290Hz
(e) Case 5, Vo = 0 and Vs = 0. ; fo = fs [(V+0)/(V+0)] = 1350Hz
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