**Sound can be defined as an elastic wave
that travels in the air and in other materials in the form of a longitudinal
wave. In air, the sound velocity increases with temperature; at 20șC is 343
m/s, approximately. Among the different wave phenomena to learn by the student,
on sound waves it is very important to know about standing waves, interference
and Doppler effect, so, below is an exercise on each one of these interesting
phenomena. **

**Standing Sound Waves**

** Musical instruments are simple sources of
sound producing standing waves. The strings of an instrument can vibrate as
a whole, with nodes only in the extremes in which case the frequency of vibration
is called fundamental. The strings can vibrate at higher frequencies and those
frequencies are called harmonics, in which case there are one or more additional
nodes. Each harmonic frequency is a multiple of the fundamental.**

**Standing Sound Wave in a String. Example.-
** ** A guitar string is 64 cm long
and has a fundamental Mi frequency of 330 Hz. When pressing in the first fret
(nearest to the tuning keys) the string is shortened in such a way that it
plays a Fa note having a frequency of 350 Hz. Calculate the distance between
this first fret and the nut necessary to get this effect. **

**Solution:
We know the total string length must be half of the fundamental wave length
L**

**The sound propagation velocity in this string
is then
v = _{1}f_{1}
= (1.28 m)(330 Hz) = 422 m/s.**

**When pressing the string, the string tension
do not varies, and the propagation velocity v is also 422 m/s.**

**The new string length L _{2} = is
obtained using
L_{2 =}
_{2}/2,
_{2}
= v/f_{2} **

**Then L _{2} = v/(2f_{2})
= (422 m/s)/(2·350 Hz) = 0.603 m. **

**The difference 64 cm - 60.3 cm = 3,7 cm
is the distance from the first fret to the nut.**

**For more examples, see links below.**

**Sound Waves Interference**

** When two waves meet at the same point in
the space at the same time it occurs the phenomena called interference. The
resulting displacement is the sum of the individual displacements of each
wave.**

**Sound Waves Interference, Example.-**
** Assume you have two loudspeakers separated
1 meter excited by the same oscillator emitting a 1150 Hz sound frequency.
You are 4 m from one of the loudspeakers. At what distance from you should
be the second loudspeaker to produce destructive interference? Assume the
air velocity is 343 m/s. **

**Solution:
The wavelength of this sound is
= v/f = (343 m/s)/(1150 Hz) = 0.3 m. To produce destructive interference,
you should be half wavelength or 0.15 cm apart from one loudspeaker than the
other. So, you should be at 4.15 or at 3.85 m from the second loudspeaker.**

**For more examples, see links below.**

**Doppler effect**

** The Doppler effect is the frequency change
of a sound perceived by the observer due to the movement of the sound source
and/or the movement of the observer. If the source and the observer approach
each other the perceived frequency increases and if they move away the perceived
frequency decreases.**

**The new frequency
detected by the observer is given by the formula
= [(v ± v**

**Problem examples on this Doppler Effect
are included in the links immediately below:**

**Sound
Waves, Doppler Effect - Examples
Sound
Waves: Standing Wave, Interference, Doppler Effect - Examples**

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