Introduction to Sound: Acoustics for the Hearing and Speech SciencesThis powerful text continues with the same fundamental aim as the previous two editions to not present, but to teach the fundamental concepts of acoustics to students in the speech-language and hearing science. In addition to relocating revised practice problems and answers to a newly created Course Notes and Workbook for Introduction to Sound, the third edition also introduces new features. Supplements Course Notes and Workshop for Introduction to Sound 1-56593-980-8 - 176 pages, softcover, 8 1/2 x 11 |
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Page 114
... problem in the same conceptual manner . Thus , 4.3010 -4.0000 + 0.3010 -3.6990 . In other words , 4.3010 is really 3.6990 units below zero , not 4.3010 units below zero . Fortunately , when you solve the problem log10 0.0002 with a ...
... problem in the same conceptual manner . Thus , 4.3010 -4.0000 + 0.3010 -3.6990 . In other words , 4.3010 is really 3.6990 units below zero , not 4.3010 units below zero . Fortunately , when you solve the problem log10 0.0002 with a ...
Page 130
... Problem 3 : What is the intensity level re : 10-12 watt / m2 of a sound with an absolute intensity of 10-6 watt / m2 ? 1. Follow the same steps that were used with the first two problems , but note that the ratio is now different . The ...
... Problem 3 : What is the intensity level re : 10-12 watt / m2 of a sound with an absolute intensity of 10-6 watt / m2 ? 1. Follow the same steps that were used with the first two problems , but note that the ratio is now different . The ...
Page 131
... Problem 4 , which is 76 dB . 2. If we solve the problem step - by - step without taking answers to previous problems into account , we must get the same result : dB = 10 log ( 4 × 10-5 ) 10-12 10 log ( 4 × 107 ) = 10 × 7.6 = = 76 . Problem ...
... Problem 4 , which is 76 dB . 2. If we solve the problem step - by - step without taking answers to previous problems into account , we must get the same result : dB = 10 log ( 4 × 10-5 ) 10-12 10 log ( 4 × 107 ) = 10 × 7.6 = = 76 . Problem ...
Contents
The Nature | 2 |
Sound Source Acting on a Medium | 10 |
The Nature of Sound Waves | 11 |
Copyright | |
12 other sections not shown
Common terms and phrases
acoustic amount amplitude angle answer antilog applied bandwidth base calculate called Chapter complex components compression concept constant corresponds curve cycle dB SPL decibels decreases defined density described determine direction displacement distance distortion elasticity energy equal Equation equilibrium example expressed factor Figure filter force fork four frequency function fundamental greater harmonic impedance increases input instantaneous intensity inversely learned length magnitude mass maximal maximum mean measure medium meter molecules motion move movement multiples occurs opposite output panel particle pendulum period phase position problem produces propagation proportional quantity ratio reactance reference reflected relation relative resistance resonant result scale shown shows signal simple sine wave sinusoidal solve sound pressure sound wave spectrum square starting string surface Table tion tube tuning unit velocity vibration waveform wheel white noise zero