16 SOUND AND HEARING 16 1 16 2 16 3 A 2 p 5 8 A 1 A 2 30 Pa 2 1 2 10 m 1 2 10 m 3 0 10 Pa p 344 m s 1000 Hz 0 344 m p max2 v f and Bk is constant gives max1 1 A IDENTIFY and SET UP Eq 15 1 gives the wavelength in terms of the frequency Use Eq 16 5 to relate the pressure and displacement amplitudes EXECUTE a b maxp BkA p max2 p max1 BkA c max p p BA p 2 max2 2 max p 3 0 10 Pa max1 p 1 5 10 Pa max2 344 m s 6 9 m 50 Hz f EVALUATE The pressure amplitude and displacement amplitude are directly proportional For the same displacement amplitude the pressure amplitude decreases when the frequency decreases and the wavelength increases IDENTIFY Apply maxp SET UP v and BA 2 so max1 1 constant 2 2 1 v and solve for A BkA f 0 344 m 6 9 m and and so k 3 p 2 p v max Bf 2 2 f v k 2 fBA 2 v 3 0 10 Pa 1480 m s 2 2 2 10 Pa 1000 Hz 9 3 21 10 m 12 EXECUTE A v B is a lot smaller and 2 and k EVALUATE Both v and B are larger but B is larger by a much greater factor so therefore A is a lot smaller IDENTIFY Use Eq 16 5 to relate the pressure and displacement amplitudes B SET UP As stated in Example 16 1 the adiabatic bulk modulus for air is to calculate from f and then f EXECUTE a 150 Hz v f Need to calculate k p max of 30 Pa b f is larger by a factor of 10 so factor of 10 max c There is again an increase in f k and maxp threshold EVALUATE When f increases decreases so k increases and the pressure amplitude increases 2 rad 150 Hz 344 m s 1 42 10 Pa 2 74 rad m 0 0200 10 m 7 78 Pa is larger by a factor of 10 and maxp of a factor of 10 so max above the pain threshold 1 42 10 Pa f v 2 3 k f v 77 8 Pa 778 Pa BkA BkA 2 so p p 2 k k 5 5 is larger by a far above the pain This is below the pain threshold 2 74 rad m Then Use Eq 15 1 Copyright 2012 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 16 1 16 2 Chapter 16 16 4 p IDENTIFY Apply max SET UP 344 m s vp max BA 2 v EXECUTE f BkA k 2 f 2 v so max p fBA 2 v 344 m s 10 0 Pa 2 1 42 10 Pa 1 00 10 m 6 5 3 86 10 Hz 3 EVALUATE Audible frequencies range from about 20 Hz to about 20 000 Hz so this frequency is audible IDENTIFY and SET UP Use the relation to find the wavelength or frequency of various sounds v f 16 5 EXECUTE a 90 m v f 1531 m s 17 Hz 1531 m s 102 kHz 0 015 m 344 m s 25 10 Hz 1 4 cm 78 kHz 3 b c v f v f f d For v f 344 m s 78 10 Hz 3 The range of wavelengths is 4 4 mm to 8 8 mm e 0 25 mm so v f 1550 m s 3 0 25 10 m 6 2 MHz 4 4 mm For f 39 kHz v f 344 m s 39 10 Hz 3 8 8 mm 16 6 16 7 v v 4 f 8 m 400 Hz 1300 kg m 1 33 10 Pa Apply Eq 16 7 for the waves in the liquid and Eq 16 8 for the waves in the 1 50 m 3 90 10 s f so EVALUATE Nonaudible to human sounds cover a wide range of frequencies and wavelengths IDENTIFY metal bar SET UP In part b the wave speed is d t EXECUTE a Using Eq 16 7 B v B 3 L t 2 2 Y v b Using Eq 16 8 1 50 m 3 90 10 s 6400 kg m 9 47 10 Pa v EVALUATE In the liquid and in the metal 3200 m s greater than the speed of sound in air IDENTIFY d v SET UP Use water EXECUTE Since along the path to the diver the sound travels 1 2 m in air the sound wave travels in water for the same time as the wave travels a distance 22 0 m 1 20 m 20 8 m in air The depth of the diver is Both these speeds are much vt for the sound waves in air and in water as given in Table 16 1 In air at 20 C 4 v 3850 m s 1482 m s 344 m s 10 v 10 2 3 2 2 2 20 8 m 20 8 m v water v air 1482 m s 344 m s 89 6 m This is the depth of the diver the distance from the horn is 90 8 m EVALUATE The time it takes the sound to travel from the horn to the person on shore is t The time it takes the sound to travel from the horn to the diver is 1 22 0 m 0 0640 s 344 m s 1 2 m 344 m s 89 6 m 0 0035 s 1482 m s t 0 0605 s 0 0640 s These times are indeed the same For three 16 8 2 figure accuracy the distance of the horn above the water can t be neglected IDENTIFY Apply Eq 16 10 to each gas SET UP In each case express M in units of kg mol For 1 41 8 3145 J mol K 300 15 K 1 41 2H EXECUTE a v H 2 2 02 10 kg mol 3 1 32 10 m s 3 For He and Ar 1 67 Copyright 2012 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher Sound and Hearing 16 3 1 67 8 3145 J mol K 300 15 K 1 02 10 m s 3 b v He c Ar v 3 4 00 10 kg mol 39 9 10 kg mol 3 1 67 8 3145 J mol K 300 15 K 323 m s T d Repeating the calculation of Example 16 4 at v v 0 928 air H EVALUATE v is larger for gases with …