SOUND WAVES EXAMPLES

Exercise 1: Doppler Effect.
The Doppler effect is used with ultrasonic waves of frequency f_s = 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 f_s = 2.25x10⁶ Hz, f^''- f_s = 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 f_s given by f^' = [(v+v₀)/v]f_s.
The heart "reflects" a new sound of frequency f^'' = [(v/(v-v₀)]f^'.
Replacing in f^''-f_s = 600 Hz results:
f^''-f_s = [(v/(v-v₀)][(v+v₀)/v]f_s - f_s = 600, where v₀ is the asked heart's velocity, v is the sound velocity in the tissues and f_s 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

For Sound Waves Theory and
More Examples, Visit:
Sound Waves Concepts and Examples
Sound Waves, Doppler Shift Effects Example

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