Complete question is;
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:
A) The level of sound in the quiet room will be 20 dB
B) The intensity of sound in the busy street is 10⁻⁵ W·m⁻²
Step-by-step explanation:
Formula given is; β = 10log(I/I₀)
(a) For Quiet room:
We are given;
I = 10⁻¹⁰ W·m⁻²
I₀ = 1 × 10⁻¹² W·m⁻²
Plugging these values into the given equation, we have;
β = 10log[(10⁻¹⁰/(1 × 10⁻¹²)]
β = 10log(10²)
β = 10 × 2 = 20 dB
Thus, the level of sound in the quiet room will be 20 dB.
(b) For the Street;
We are given;
β(street) - β(room) = 50 dB
Now, let's rewrite the given intensity level equation;
β = 10logI - 10 logI₀
Now, Let the intensity level for the room be β₁ and let the intensity level for the road be β₂. Thus;
β₁ = 10logI₁ - 10log I₀ - - - - (eq 1)
β₂ = 10logI₂ - 10logI₀ - - - - (eq 2)
Subtract eq 1 from eq 2 to give;
β₂ - β₁ = 10logI₂ - 10logI₁
50 = 10logI₂ - 10log(10⁻¹⁰)
Divide each term by 10 to give:
5 = logI₂ - log(10⁻¹⁰)
5 = logI₂ - (-10)
5 = logI₂ + 10
Subtract 10 from each side to give;
-5 = logI₂
Taking the antilog of both sides to give;
I₂ = 10⁻⁵ W·m⁻²
Thus, the intensity of sound in the busy street is 10⁻⁵ W·m⁻².
