A loud speaker at a rock concert generates 102 W/m2 at 20m with a frequency of 1
kHz. Assume that the speaker spreads its intensity uniformly in all direction.
(a) What is the sound level ( = ?) at 20 m?
(b) What is the total acoustic power output of the speaker?
(c) At what distance will intensity level be at the pain threshold of 120 dB?
(d) Determine the sound intensity at a distance of 30 m?
To answer these questions, we'll need to use the formulas and equations related to sound intensity and sound level. Let's go through each question one by one.
(a) To find the sound level (β) at 20 m, we can use the formula:
β = 10 * log10(I / I₀)
where I is the sound intensity and I₀ is the reference intensity, typically taken as 10^(-12) W/m².
In this case, we are given the sound intensity at 20 m as 10^(-2) W/m². So the equation becomes:
β = 10 * log10(10^(-2) / 10^(-12))
Simplifying this equation, we get:
β = 10 * log10(10^(10))
β = 10 * 10
β = 100 dB
Therefore, the sound level at 20 m is 100 dB.
(b) The total acoustic power output of the speaker can be calculated using the equation:
Power = Intensity * Area
Since the speaker spreads its intensity uniformly in all directions, the intensity at any distance is the same. Therefore, we can use the given intensity at 20 m and multiply it with the surface area of a sphere with a radius of 20 m.
Surface area of a sphere = 4πr²
= 4 * 3.14 * (20)^2
= 5024 m²
Power = 10^(-2) W/m² * 5024 m²
= 50.24 W
Therefore, the total acoustic power output of the speaker is 50.24 W.
(c) To find the distance at which the intensity level reaches the pain threshold of 120 dB, we can rearrange the formula for sound level:
β = 10 * log10(I / I₀)
to solve for distance (r):
r = sqrt(Power / (4πI₀ * 10^((β-120)/10)))
Here, Power is the total acoustic power output of the speaker, given as 50.24 W, and β is the desired sound level of 120 dB.
Plug in these values and solve for r:
r = sqrt(50.24 / (4π * 10^(-12) * 10^((120-120)/10)))
r = sqrt(50.24 / (4π * 10^(-12)))
r ≈ 70834 m
Therefore, at a distance of approximately 70834 m, the intensity level will be at the pain threshold of 120 dB.
(d) To determine the sound intensity at a distance of 30 m, we can use the inverse square law:
I₂ = I₁ * (r₁/r₂)^2
Where I₁ is the sound intensity at a reference distance of r₁, and I₂ is the sound intensity at a different distance r₂.
Given that I₁ is 10^(-2) W/m² at 20 m, we can use this formula to find I₂ at 30 m:
I₂ = 10^(-2) W/m² * (20/30)^2
I₂ = 10^(-2) * (2/3)^2
I₂ ≈ 4.44 * 10^(-3) W/m²
Therefore, at a distance of 30 m, the sound intensity is approximately 4.44 * 10^(-3) W/m².