SOUND WAVES EXAMPLES
Exercise 1: Doppler Effect.
The Doppler effect is used with ultrasonic waves of frequency fs = 2.25x106 Hz to watch the beat of a fetus's heart. A maximum pulsation frequency of 600 Hz is detected. Assuming the sound velocity in the tissues is v = 1.54x103 m/s, calculate the maximum velocity v0 on the surface of the beating heart.
Solution: A pulsation occurs when two waves of similar frequency meet at the same point. In this case the 600 Hz pulsation is the difference between the received wave frequency f'' from the heart's surface and the original frequency fs = 2.25x106 Hz, f''- fs = 600 Hz
To calculate the received wave frequency f''
we must realize that there are two Doppler shifts that generate f'':
The heart, moving with velocity v0 "detects" a wave with a frequency f' slightly greater than fs given by f' = [(v+v0)/v]fs.
The heart "reflects" a new sound of frequency f'' = [(v/(v-v0)]f'.
Replacing in f''-fs = 600 Hz results:
f''-fs = [(v/(v-v0)][(v+v0)/v]fs - fs = 600, where v0 is the asked heart's velocity, v is the sound velocity in the tissues and fs is the ultrasonic source frequency.
Replacing values and solving for v0 it is finally obtained v0 = 0.205 m/s.
Exercise 2: Stationary Sound Waves.
A guitar string has a total length of 90 cm and a mass of 3.6 g. From the bridge to the nut there is a distance of 60 cm and the string has a tension of 520 N. Calculate the fundamental frequency and the first two overtones.
Solution: The relation between string tension T, mass m, length L and string wave velocity v is
T = (m/L)·v2
(m/L) is the string mass per unit length. In this case m/L = 3.6x10-3
kg/0.9m = 4x10-3 kg/m.
Replacing T = 520 N and m/L = 4x10-3
kg/m, we get v = 361 m/s.
The wavelength associated to the fundamental frequency is 2·(string length) = 2·0.6 m = 1.2 m. The fundamental frequency is then v/ = (361m/s)/1.2m = 301 Hz.
The first and second overtones are respectively 602 Hz and 903 Hz.
Exercise 3: Interference Sound Waves.
Two loudspeakers are separated by 2.5 m. A person stands at 3.0 m from one and at 3.5 m from the other one. Assume a sound velocity of 343 m/s.
a) What is the minimum frequency to present destructive interference at this point?
b) Calculate the other two frequencies that also produce destructive interference.
Solution: To get destructive interference
the difference between the distances to the loudspeakers should be n/2,
n = 1, 3, 5 ... There will be destructive interference at /2,
at 3/2 and at
5/2. As the
difference in distance is 3.5 m - 3.0 m = 0.5m, then for destructive interference
= 1.0 m and
f1 = v/ = (343
m/s)/1.0 m = 343 Hz.
The wavelength of the next frequency that also produces destructive interference is obtained doing 3/2 = 0.5 m or = 1/3 m and then f2 = (343 m/s)/(1/3 m) = 1029 Hz.
Similarly, doing 5/2 = 0.5m or = 1/5 m we get f3 = 1715 Hz
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