A description of how and why sound intensity decreases with distance with a short practice problem

1. Intensityisrelatedtothe loudnessof asound. Loudnessis alsorelatedto the soundpower,or rate of
energytransferintosoundenergy. The powersendssoundwavesoutuniformlyfromsource inevery
direction,likeagrowingsphere.
Red – soundsource
Blue – sphere showinghow
powertravelsaway
from the source over
time
time
As the powertravelsfartherandfartherfromthe source,the hypothetical sphereincreasesinsize.Since
intensityequalspowerdividedbyarea, =
𝑃
𝐴
,in 3 dimensionsintensityequals powerdividedbysurface
area of a sphere, 𝐼 =
𝑃
4𝜋𝑟2
.The surface area the powerisspreadoverisincreasing;therefore,the
intensitydecreaseswithincreasingdistance.The loudnessof the soundwe hearalsodecreases.
PRACTICEPROBLEM
You are walkingtoschool andaccidentallydropyourfriend’scellphoneonthe ground. Luckily,itisnot
damaged,butthe collisionwiththe groundproduces50dBof soundfrom 1m away.If the path to school
isstraightand you knowyourfriend isonthe same pathbehindyou,how farwouldyourfriendhave to
be to not hearany trace of the accident?Assume there are noothersoundsandthe soundof the
collisionisonlyaffectedbydistance.
CRASH!

2. SOLUTION
Firstwe needto change decibelstowatts permetersquared
𝐵 = 0𝑑𝐵 + 10 log (
𝐼
𝐼𝑜
)
50𝑑𝐵 = 0𝑑𝐵 + 10 log(
𝐼
(10−12) 𝑊𝑚2
)
105 = (
𝐼
(10−12) 𝑊𝑚2
)
𝐼 = 1.0 × 10−6 𝑊𝑚2
Findthe powerof the soundusingthe intensityfrom1 meteraway:
𝐼 =
𝑃
4𝜋𝑟2
1.0 × 10−6𝑊𝑚2
=
𝑃
4𝜋
𝑃 = 1.26 × 10−5 𝑊
Minimumintensityof whathumanscan detect: 10−12 𝑊𝑚2 *assume thisisthe absolute minimum
intensityyourfriendcandetect
𝐼 =
𝑃
4𝜋𝑟2
10−12 𝑊𝑚2 =
1.26×10−5 𝑊
4𝜋𝑟2
Solve forr: r isthe distance fromthe 1 meterradiusof the collisionthatthe sound intensityshouldfade
to the veryminimumlevel humanscandetect.
𝑟 = 1000𝑚
You friendmustbe 1001m behindyou