155k views
0 votes
You and a friend are watching fireworks that are launching from Toledo Volleyball Court. You are standing at Hensville Park (570 m) and your friend is at the Maritime Academy of Toledo (669 m). The sound intensity at the Maritime Academy is 0.2 W/m2.

What is the sound intensity level at your location?

User Lupina
by
8.2k points

1 Answer

2 votes

Final answer:

To determine the sound intensity level at Hensville Park (570 m away), we use the inverse square relationship of sound intensity with distance. By knowing the intensity at the Maritime Academy (669 m away) is 0.2 W/m², we can calculate that the intensity at Hensville Park is approximately 0.2755 W/m². Then, we can calculate the sound intensity level in decibels using the standard formula.

Step-by-step explanation:

The question asks to determine the sound intensity level at a location that is 570 meters away from the source of the fireworks, given that the sound intensity is 0.2 W/m² at another location that is 669 meters away. The sound intensity I is related to the distance r from the source in an inverse square relationship, given by the formula I = P / (4πr²), where P is the power of the sound source.

For two different distances, r1 and r2, from the same sound source, this relationship implies that I1 / I2 = (r2²) / (r1²). Using the distance r1 = 669 m and r2 = 570 m and the intensity I2 = 0.2 W/m² at r1, we can calculate the intensity I1 at r2.


Since I1 / 0.2 W/m² = (669 m / 570 m)²,
then I1 = 0.2 W/m² × (669 m / 570 m)².

I1 = 0.2 W/m² × (1.1737)²
I1 = 0.2 W/m² × 1.3776
I1 ≈ 0.2755 W/m²

The sound intensity level L in decibels (dB) can be found using the formula L = 10 × log10(I / I0), where I0 = 1× 10⁻¹² W/m² is the reference intensity level. Substituting the calculated intensity gives the intensity level at the 570 m location.

User Kevin Donde
by
7.8k points