(a) Explanation Given info: The intensity level of sound in the church is 120 dB and the distance is 3.00 m . The reference intensity of the sound is 1.00 × 10 − 12 W / m 2 . Write the expression for the intensity level of the sound. β = 10 log ( I I 0 ) (1) Here, I is the intensity of the sound. I 0 is the reference intensity. For sound level 120 dB : Substitute 1.00 × 10 − 12 W / m 2 for I o and 120 dB for β in above equation (1). 120 dB = 10 log ( I 1 1.00 × 10 − 12 W / m 2 ) I 1 = 1 W / m 2 Thus the intensity of the sound at 3.00 m distance is 1 W / m 2 . For sound level 100 dB : Substitute 1.00 × 10 − 12 W / m 2 for I o and 100 dB for β in above equation (1). 100 dB = 10 log ( I 2 1.00 × 10 − 12 W / m 2 ) I 2 = 0 (b) To determine The distance at which the sound level is 10 dB .

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ISBN: 9780100461260

Author: SERWAY

Publisher: YUZU

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Chapter 17, Problem 17.35P

(a)

To determine

The distance at which the sound level is 100 dB .

(b)

To determine

The distance at which the sound level is 10 dB .

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Chapter 17, Problem 17.34P

Chapter 17, Problem 17.36P

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Chapter 17 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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The distance at which the sound level is 100 dB .

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The distance at which the sound level is 100 dB .

The distance at which the sound level is 10 dB .

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Chapter 17 Solutions