**SOUND WAVES EXAMPLES**

**Exercise 1: Doppler Effect.**

The Doppler effect is used with ultrasonic waves
of frequency f_{s} = 2.25x10^{6} 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 = 1.54x10^{3} m/s, calculate
the maximum velocity v_{0} 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} = 2.25x10^{6}
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 v_{0} "detects" a wave with a frequency
f^{'} slightly greater than f_{s} given by f^{'} =
[(v+v_{0})/v]f_{s}.

The heart "reflects" a new sound of frequency f^{''} = [(v/(v-v_{0})]f^{'}.

Replacing in f^{''}-f_{s} = 600 Hz results:

f^{''}-f_{s} = [(v/(v-v_{0})][(v+v_{0})/v]f_{s} - f_{s} = 600, where v_{0} 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 v_{0} it is finally obtained v_{0} = 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)·v^{2}

(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 f_{2} = (343 m/s)/(1/3 m) = 1029 Hz.

Similarly, doing 5/2
= 0.5m or =
1/5 m we get f_{3} = 1715 Hz

**For Sound Waves Theory and More Examples,
Visit:**

Sound Waves, Doppler Shift Effects Example

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