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 ℃ respectively).

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  1. Calculate the length of the 16 cycle wave train.
  2. 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: MacIsaac, Dan and Ari Hämäläinen. "Physics and technical characteristics of ultrasonic sonar systems." The Physics Teacher. Vol. 40, No. 1 (January 2002): 39-46.

solution

Solutions …

  1. There are two ways to do this part of the problem. One is to calculate one wavelength …
    λ =  v  =  343 m/s  = 0.00694 m = 6.94 mm
    f 49,400 Hz
    and then multiply by 16 to get the length of the whole train.
     
    s = 6.94 mm × 16 = 111 mm
     
    The other is to determine the time for 16 pulses …
     
    t = 16/49,400 Hz = 3.24 × 10−4 s
     
    and then multiply by speed to get distance traveled.
     
    s = vt = (343 m/s)(3.24 × 10−4 s) = 111 mm
     
  2. We've already finished computing the first entry while working on part a. Two decimal places should provide adequate precision.
     
    t = 16/49,400 Hz = 3.24 × 10−4 s = 0.32 ms
     
    Next we'll do the last entry. The time for one complete cycle is one period, which is the inverse of frequency.
             
    T =  1  =  1  = 0.020 s = 20.0 ms
    f 50 Hz
             
    The listening phase is the remaining time.
     
    t = 20.0 ms − (2.38 ms + 0.32 ms) = 17.3 ms
     
    To compute round trip distances, multiply speed by round trip time.
         
    srt = vtrt   srt = vtrt
    srt = (343 m/s)(2.38 ms)   srt = (343 m/s)(17.3 ms)
    srt = 816 mm   srt = 5930 mm
         
    Divide the round trip distances by 2 to get the one way distances.
         
    s = srt/2   s = srt/2
    s = (816 mm)/2   s = (5930 mm)/2
    s = 408 mm   s = 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 — 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 — 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.

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