SME
University Physics III: Waves, Optics & Modern · Problem type
Sound intensity and decibels at a distance
Use this for “A loudspeaker sends out 2.5 W of sound evenly in all directions. What are the intensity and the sound level 15 m away?”
Table
Distance (r)
(m)
Sound level (β)
(dB)
This is an example. Type your own values to start a new problem.
I (W/m²)
=
P (
)
4π
(
r (
)
)
2
β
Sound level
calculated · dB
Formulas
I
=
P
4π ×
r
2
7.9577 ×
10
-4
=
1
4π ×
10
2
β
=
10
×
log₁₀(
I
10
-12
)
89.0079
=
10
×
log₁₀(
7.9577 ×
10
-4
10
-12
)
Assumptions
•
The source sends its power evenly in every direction, and nothing absorbs or reflects it, so all of P crosses a sphere of area 4πr².
•
The sound level compares I with the threshold of hearing I₀ = 10⁻¹² W/m²: β = 10 log₁₀(I ÷ I₀) in decibels.
•
Twice the distance spreads the power over four times the area: a quarter of the intensity, about 6 dB lower.
Step-by-step
WE KNOW
P = 1 W, r = 10 m
FIND
intensity (I), sound level (β)
Step 1 · Find intensity (I)
Use I = P ÷ (4π × r²)
The
power
spreads
over
a
sphere
of
radius
r
:
divide
it
by
the
sphere’s
area
4πr².
I
=
1
4π ×
10
2
I
=
7.9577 ×
10
-4
W/
m
2
Step 2 · Find sound level (β)
Use β = 10 × log₁₀(I ÷ 10⁻¹²)
Count the powers of ten above the threshold I₀, 10 dB for each.
β
=
10
×
log₁₀(
7.9577 ×
10
-4
10
-12
)
β
=
10
×
log₁₀(
7.9577 ×
10
-4
10
-12
)
β
=
10
×
log₁₀(7.9577
×
10
8
)
β
=
10
×
8.901
β
=
89.0079
dB
Check
Put the numbers back in. Both sides should match.
7.9577 ×
10
-4
=
1
4π ×
10
2
✓
89.0079
=
10
×
log₁₀(
7.9577 ×
10
-4
10
-12
)
✓
Topic check
See what you know: 5 questions on this topic
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