To determine whether a small speaker delivering 2 W of audio can produce a sound intensity of 120 dB, we can follow these steps:
### Step 1: Understand the relationship between intensity and decibels
The loudness of sound in decibels (dB) is given by the formula:
\[
L = 10 \log \left( \frac{I}{I_0} \right)
\]
where:
- \(L\) is the sound level in decibels,
- \(I\) is the intensity of the sound in watts per meter squared (W/m²),
- \(I_0\) is the reference intensity, which is \(10^{-12} \, \text{W/m}^2\).
### Step 2: Set up the equation with known values
We know that:
- \(L = 120 \, \text{dB}\)
- \(I_0 = 10^{-12} \, \text{W/m}^2\)
Substituting these values into the formula:
\[
120 = 10 \log \left( \frac{I}{10^{-12}} \right)
\]
### Step 3: Solve for intensity \(I\)
Divide both sides by 10:
\[
12 = \log \left( \frac{I}{10^{-12}} \right)
\]
Now, remove the logarithm by rewriting it in exponential form:
\[
10^{12} = \frac{I}{10^{-12}}
\]
Multiplying both sides by \(10^{-12}\):
\[
I = 10^{12} \times 10^{-12} = 1 \, \text{W/m}^2
\]
### Step 4: Relate intensity to power and distance
The intensity \(I\) is also related to the power \(P\) of the speaker and the distance \(r\) from the speaker:
\[
I = \frac{P}{A} = \frac{P}{4 \pi r^2}
\]
where \(A\) is the area over which the power is distributed, and for a spherical wave, \(A = 4 \pi r^2\).
### Step 5: Rearrange to find distance \(r\)
Rearranging the equation gives:
\[
r^2 = \frac{P}{4 \pi I}
\]
Taking the square root:
\[
r = \sqrt{\frac{P}{4 \pi I}}
\]
### Step 6: Substitute the known values
Substituting \(P = 2 \, \text{W}\) and \(I = 1 \, \text{W/m}^2\):
\[
r = \sqrt{\frac{2}{4 \pi \times 1}} = \sqrt{\frac{2}{4\pi}} = \sqrt{\frac{1}{2\pi}}
\]
### Step 7: Calculate the distance
Calculating the value:
\[
r \approx \sqrt{\frac{1}{6.2832}} \approx \sqrt{0.15915} \approx 0.399 \, \text{m}
\]
### Conclusion
The distance \(r\) is approximately \(0.399 \, \text{m}\) or about \(39.9 \, \text{cm}\). Therefore, at this distance, the sound intensity will be \(120 \, \text{dB}\).
