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.
We know the total string length must be half of the fundamental wave length
L1 = 0.64 m = 1/2 or 1 = 1.28 m
The sound propagation velocity in this string
v = 1f1 = (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 L2 = is
L2 = 2/2, 2 = v/f2
Then L2 = v/(2f2) = (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.
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.
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 ± vo )/(v vs )] · f
where v is the propagation sound velocity, v0 is the observer velocity and vs is the source velocity . The upper signs (+ and -) apply if the source and/or the observer approach, and the lower signs (- and +) if they separate.
Problem examples on this Doppler Effect are included in the links immediately below:
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