$L =$ loudness, in decibels $( dB ) ; I =$ sound intensity, in watts $/ m ^2 ; I_0=10^{-12}$ watts $/ m ^2$
The loudness of a jack hammer is 96 dB . Its sound intensity is about $\square$ 0.004
The loudness of a compactor is 94 dB . Its sound intensity is about
Answer (2)
Using the formula for loudness in decibels, we find that the sound intensity of the compactor, which has a loudness of 94 dB, is approximately 0.00251 watts/m 2 . This was calculated by applying the loudness formula to the given values. Similar calculations confirm the sound intensity of the jack hammer at 96 dB is about 0.004 watts/m 2 .
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Use the formula L = 10 lo g 10 ( I 0 I ) to relate loudness and sound intensity.
For the jack hammer, 96 = 10 lo g 10 ( 1 0 − 12 I ja c khamm er ) . Solve for I ja c khamm er ≈ 0.00398 watts/m 2 .
For the compactor, 94 = 10 lo g 10 ( 1 0 − 12 I co m p a c t or ) . Solve for I co m p a c t or ≈ 0.00251 watts/m 2 .
The sound intensity of the compactor is 0.00251 watts/m 2 .
Explanation
Understanding the Problem We are given the formula for loudness L in decibels (dB) as a function of sound intensity I in watts/m 2 :
L = 10 lo g 10 ( I 0 I )
where I 0 = 1 0 − 12 watts/m 2 is the reference intensity.
We are given that the loudness of a jack hammer is 96 dB and the loudness of a compactor is 94 dB. We want to find the sound intensity of the compactor.
Jack Hammer Intensity First, let's find the sound intensity of the jack hammer, I ja c khamm er . We have:
96 = 10 lo g 10 ( 1 0 − 12 I ja c khamm er )
Dividing both sides by 10, we get:
9.6 = lo g 10 ( 1 0 − 12 I ja c khamm er )
Taking the antilog (i.e., raising 10 to the power of both sides), we have:
1 0 9.6 = 1 0 − 12 I ja c khamm er
Multiplying both sides by 1 0 − 12 , we get:
I ja c khamm er = 1 0 9.6 × 1 0 − 12 = 1 0 − 2.4 ≈ 0.00398 watts/m 2
This confirms the given information that the sound intensity of the jack hammer is about 0.004 watts/m 2 .
Compactor Intensity Now, let's find the sound intensity of the compactor, I co m p a c t or . We have:
94 = 10 lo g 10 ( 1 0 − 12 I co m p a c t or )
Dividing both sides by 10, we get:
9.4 = lo g 10 ( 1 0 − 12 I co m p a c t or )
Taking the antilog (i.e., raising 10 to the power of both sides), we have:
1 0 9.4 = 1 0 − 12 I co m p a c t or
Multiplying both sides by 1 0 − 12 , we get:
I co m p a c t or = 1 0 9.4 × 1 0 − 12 = 1 0 − 2.6 ≈ 0.00251 watts/m 2
Final Answer Therefore, the sound intensity of the compactor is approximately 0.00251 watts/m 2 .
Examples
Understanding sound intensity is crucial in many real-world scenarios. For example, city planners use decibel measurements to create noise maps, helping them to mitigate noise pollution near airports or highways. Similarly, audiologists rely on these calculations to assess hearing damage risk and recommend appropriate hearing protection. By quantifying sound levels, we can better protect public health and improve environmental quality.
