Physics Bloopers: Sound Level Problem Gone Wrong

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SUMMARY

The discussion centers on a physics problem involving sound levels measured in decibels (dB). The problem states that a sound source produces 105 dB at a distance of 5.0 m, and the goal is to determine how far an observer must be to not hear the sound at all. The solution involves using the reference intensity of 1 x 10-12 W/m2 and applying the inverse square law. The initial calculation incorrectly suggested a distance of approximately 890 km, highlighting the importance of understanding the difference between relative and absolute sound levels.

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  • Familiarity with the inverse square law in physics
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In my high school physics class the other day the teacher botched an example problem involving sound level (decibels).

Homework Statement


A single sound source produces 105 dB at a distance of 5.0 m. How far away must the observer be to not hear this sound at all?


Homework Equations


beta.jpg



The Attempt at a Solution


The reference intensity I0 is the lowest intensity a human can hear, 1 x 10-12 W/m2, so when the source is far enough away its intensity will be equal to this. The intensity of a sound falls off as 1/d2, so after substituting the information in the problem the equation above can be rewritten as follows:
sub.jpg

In this equation, d0 represents the distance to the sound source that produces the reference intensity and d is the distance to the original source, 5.0 m for this problem.
When this is solved for d0, the result is 177828d, or about 890 km. This can't be right! What happened? Thanks in advance for your help.
 
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EDIT: Whoops. This is a really old question. I answered it by mistake thinking it was recent.

keydetpiper said:
In my high school physics class the other day the teacher botched an example problem involving sound level (decibels).

Homework Statement


A single sound source produces 105 dB at a distance of 5.0 m. How far away must the observer be to not hear this sound at all?

Homework Equations


View attachment 126674

The Attempt at a Solution


The reference intensity I0 is the lowest intensity a human can hear, 1 x 10-12 W/m2, so when the source is far enough away its intensity will be equal to this. The intensity of a sound falls off as 1/d2, so after substituting the information in the problem the equation above can be rewritten as follows:
View attachment 126675
In this equation, d0 represents the distance to the sound source that produces the reference intensity and d is the distance to the original source, 5.0 m for this problem.
When this is solved for d0, the result is 177828d, or about 890 km. This can't be right! What happened? Thanks in advance for your help.
$$\text{β = 10log(}\frac{d₀²}{d²})$$ is unhelpful in this question.

The formula only applies to relative (not absolute) dB values. E.g. if d₀ = 10d, then β = 10log(10²) = 20dB.

That means there is a 20dB *difference* between levels at the two distances. The absolute levels (referenced to 10⁻¹²W/m²) could be 100dB and 80dB, or 20dB and 0dB for example, depending on ther source's power output.

The formula contains no information about the absolute threshold level (10⁻¹²W/m²). For example if the threshold level were changed to, say, 10⁻¹¹W/m², the formula would give the same result, which makes no sense.

I’d take this approach:

Use 105 = 10log(I/10⁻¹²) to work out I/10⁻¹².

Apply the inverse square law:$$\frac{d₀²}{5²} = \frac{I}{10⁻¹²}$$to find d₀.
 
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