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Use information from Table 16.2 to answer the following

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 14E Chapter 16

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 14E

Use information from Table 16.2 to answer the following questions about sound in air. At 20°C the bulk modulus for air is 1.42 × 105 Pa and its density is 1.20 kg/m3. At this temperature, what are the pressure amplitude (in Pa and atm) and the displacement amplitude (in m and nm) (a) for the softest sound a person can normally hear at 1000 Hz and (b) for the sound from a riveter at the same frequency? (c) How much energy per second does each wave deliver to a square 5.00 mm on a side?

Step-by-Step Solution:

Solution 14E Step 1 of 8: (a) or the softest sound a person can normally hear at 1000 Hz and (b) Sound waves may also be described in terms of variations of pressure at various points. In a sinusoidal sound wave in air, the pressure fluctuates above and below atmospheric pressure in a sinusoidal variation with the same frequency as the motions of the air particles. Given data, 5 Bulk modulus of air, B=1.42× 10 pa Density of air, =1.20 kg/m3 Frequency of sound, f= 1000 Hz From table Intensity, I=1012W/m 2 Step 2 of 8: To find, Speed of sound wave, v= Pressure amplitude, P= Displacement amplitude, A= To calculate the speed of sound, Using B v = Substituting B=1.42× 10 pa and =1.20 kg/m 3 1.42×10 pa v = 1.20 kg/m v =344 m/s Therefore, the speed of sound in air is 344 m/s. Step 3 of 8: To calculate the pressure amplitude, Using I = P2 2v Solving for pressure P = 2vI Substituting =1.20 kg/m , v=344 m/s and I=10 1W/m 2 P = 2(1.20 kg/m )(344 m/s)(1012W/m )2 6 P= 28.69× 10 pa 6 Using 1 pa =9.8 × 10 atm P=2.8 × 10 atm Step 4 of 8: To calculate the Displacement amplitude, Using P= BkA Solving for amplitude, P A = Bk 2f Since k = v A = Pv 2fB 6 5 Substituting P= 28.69× 10 pa , v= 344 m/s , f= 1000 Hz and B=1.42× 10 pa 6 A = (28.69×10 pa)(344 m/s) 2(3.14)(1000)(1.42) 9 A=1106× 10 m 9 Using 1nm = 1 × 10 m A=1106nm

Step 5 of 8

Chapter 16, Problem 14E is Solved
Step 6 of 8

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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Use information from Table 16.2 to answer the following

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