Question

The decibel level of sound is 50 dB greater on a busy street than in a quiet room where the intensity of sound is 10^-10 watt/m2. The level of sound in the quiet room is (10,20,100) dB, and the intensity of sound in the busy street is (10^-1, 10^-5, 10^-10) watt/m2.

Use the formula , β = 10log I/I 0 where β is the sound level in decibels, I is the intensity of sound you are measuring, and Io is the smallest sound intensity that can be heard by the human ear (roughly equal to 1 x 10^-12 watts/m2).

Answer 1

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Explanation:

Answer 2

Answers:

(a) 20 dB; (b) 10⁻⁵ W·m⁻²

Explanation:

Step-by-step explanation:

β = 10log(I/I₀)

(a) Quiet room:

Data:

I = 10⁻¹⁰ W·m⁻²

I₀ = 1 × 10⁻¹² W·m⁻²

Calculation:

β = 10log[(10⁻¹⁰/(1 × 10⁻¹²)] = 10log(10²) = 10 × 2 = 20 dB

The level of sound in the quiet room is 20 dB.

(b) Street

Data:

β(street) - β(room) = 50 dB

Calculations:

Let's rewrite the intensity level equation as

β = 10logI - 10 logI₀

Let 1 = the room and 2 = the road. Then,

(1) β₂ = 10logI₂ - 10logI₀

(2) β₁ = 10logI₁ - 10log I₀

Subtract (2) from (1) β₂ - β₁ = 10logI₂ - 10logI₁

50 = 10logI₂ - 10log(10⁻¹⁰)

Divide each side by 10 5 = logI₂ - log(10⁻¹⁰)

5 = logI₂ - (-10)

5 = logI₂ + 10

Subtract 10 from each side -5 = logI₂

Take the antilog of each side I₂ = 10⁻⁵ W·m⁻²

The intensity of sound in the busy street is 10⁻⁵ W·m⁻².