Question

After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. The rockets explode high in the air and the sound travels out uniformly in all directions. If the sound intensity is 1.52 10-6 W/m2 at a distance of 123 m from the explosion, at what distance from the explosion is the sound intensity half this value

Answer 1

Answer:

The distance when the intensity is halved is 173.95 m

Explanation:

Given;

initial intensity of the sound, I₁ = 1.52 x 10⁻⁶ W/m²

initial distance from the explosion, d₁ = 123 m

final intensity of the sound, I₂ = ¹/₂ (1.52 x 10⁻⁶ W/m²) = 0.76 x 10⁻⁶ W/m²

Intensity of sound is inversely proportional to the square of distance between the source and the receiver.

I ∝ ¹/d²

I₁d₁² = I₂d²

(1.52 x 10⁻⁶)(123)² = (0.76 x 10⁻⁶)d₂²

d₂² = (1.52 x 10⁻⁶ x 123²) / (0.76 x 10⁻⁶)

d₂² = 30258

d₂ = √30258

d₂ = 173.95 m

Therefore, the distance when the intensity is halved is 173.95 m

Answer 2

Answer:

174m

Explanation:

We can calculate the power of the sound:

$P = IA = I \pi R^2 = 1.52\times10^{-6}*\pi*123^2 = 0.0722 W$

For the intensity to be half of I, then the cover area must be

$a = P/(I/2) = 2P/I = \frac{2*0.722}{1.52\times10^{-6}} = 95058 m^2$

Then the distance from the source center to this point is

$a = \pi r^2$

$r^2 = a / \pi = 95058 / \pi = 30258$

$r = \sqrt{30258} = 174 m$