The Nature of Sound

Practice

practice problem 1

A typical ultrasonic ranger found in a science classroom emits a 49.4 kHz sound wave that is pulsed 50 times a second. The ultrasound is inaudible, but the beginning of each pulse produces in an audible click. 50 clicks per second gives the ranger its characteristic buzzing sound. The transducer is driven through 16 vibrations at the beginning of each cycle. This is followed by a 2.38 ms "rest" period to allow the transducer to calm down. During the remainder of the cycle, the transducer "listens" for echoes. The circuitry attached to the transducer calculates the distance from the ranger to the source of the echo based on the round trip time and the speed of sound in air at room temperature (which are assumed to be 343 m/s and 20 °C respectively).

Calculate the length of the 16 cycle wave train.

Complete the following table. (Entries marked n/a are not applicable.)

	duration (ms)	round trip distance (mm)	distance from ranger (mm)	corresponding specification
16 λ		n/a	n/a	n/a
rest	2.38
listening
full cycle		n/a	n/a	n/a

Source: Physics and technical characteristics of ultrasonic sonar systems.

solution

There are two ways to find the length of the 16 cycle wave train. One is to calculate the length of one wave…

λ =	v
	f

λ =	343 m/s
	49,400 Hz

λ = 0.00694 m = 6.94 mm

and then multiply by 16 to get the length of the whole train.

∆s = 6.94 mm × 16
∆s = 111 mm

The other is to determine the time for 16 cycles…

∆t = 16/49,400 Hz
∆t = 3.24 × 10⁻⁴ s

and then multiply by speed to get distance traveled.

∆s = v∆t
∆s = (343 m/s)(3.24 × 10⁻⁴ s)
∆s = 111 mm

We've already computed the duration of the 16 cycle wavetrain. Let me rewrite it. Two decimal places should provide adequate precision.

∆t = 16/49,400 Hz
∆t = 3.24 × 10⁻⁴ s
∆t = 0.32 ms

Next we'll work on the full operational cycle. Period is the inverse of frequency.

T =	1
	f

T =	1
	50 Hz

T = 0.020 s = 20.0 ms

The listening phase is the total time minus the 16 cycles and the rest phase.

∆t = 20.0 ms − 0.32 ms − 2.38 ms
∆t = 17.3 ms

To compute round trip distances (s_rt), multiply speed (v) by round trip time (t_rt).

rest
∆s_rt = v∆t_rt ∆s_rt = (343 m/s)(2.38 ms) ∆s_rt = 816 mm

listening
∆s_rt = v∆t_rt ∆s_rt = (343 m/s)(17.3 ms) ∆s_rt = 5930 mm

Divide the round trip distances by 2 to get the one way distances (s_ow).

rest
∆s_ow = ½∆s_rt ∆s_ow = ½(816 mm) ∆s_ow = 408 mm

listening
∆s_ow = ½∆s_rt ∆s_ow = ½(5930 mm) ∆s_ow = 2970 mm

Since the ultrasonic ranger can't listen for an echo until it stops ringing it can't detect objects closer than about ½ meter (408 mm) — the minimum measurable distance. Since the ultrasonic ranger has to stop listening to start a new cycle of measurements it can't detect objects farther away than about 3 meters (2970 mm) — the maximum measurable distance. We may now complete the table.

	duration (ms)	round trip distance (mm)	distance from ranger (mm)	corresponding specification
16 λ	0.32	n/a	n/a	n/a
rest	2.38	816	408	minimum measurable distance
listening	17.3	5930	2970	maximum measurable distance
full cycle	20.0	n/a	n/a	n/a

practice problem 2

Write something else.

solution

Answer it.

practice problem 3

Write something different.

solution

Answer it.

practice problem 4

Write something completely different.

solution

Answer it.