The speaker of a public address system emits 20 kW power, considering it a point source. What is the sound intensity level at a point 4.00 m away?

Options

A. 100 dB
B. 140dB
C. 120dB
D. 125dB

Answer

To solve the problem of finding the sound intensity level at a point 4.00 m away from a speaker emitting 20 kW of power, we can follow these steps:

### Step 1: Understand the given data
- Power emitted by the speaker, \( P = 20 \, \text{kW} = 20 \times 10^3 \, \text{W} \)
- Distance from the speaker, \( r = 4.00 \, \text{m} \)

### Step 2: Calculate the intensity of the sound
The intensity \( I \) of sound from a point source is given by the formula:
\[ I = \frac{P}{A} \]
where \( A \) is the surface area of a sphere with radius \( r \):
\[ A = 4 \pi r^2 \]

Substituting the value of \( r \):
\[ A = 4 \pi (4)^2 = 4 \pi \times 16 = 64 \pi \, \text{m}^2 \]

Now, substituting \( A \) into the intensity formula:
\[ I = \frac{20 \times 10^3}{64 \pi} \]

### Step 3: Calculate the numerical value of intensity
Using \( \pi \approx 3.14 \):
\[ A \approx 64 \times 3.14 = 200.96 \, \text{m}^2 \]
Now, substituting this back into the intensity equation:
\[ I \approx \frac{20 \times 10^3}{200.96} \approx 99.52 \, \text{W/m}^2 \]

### Step 4: Convert intensity to sound intensity level in decibels
The sound intensity level \( L \) in decibels (dB) is given by:
\[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \]
where \( I_0 = 10^{-12} \, \text{W/m}^2 \) is the reference intensity.

Substituting the calculated intensity:
\[ L = 10 \log_{10} \left( \frac{99.52}{10^{-12}} \right) = 10 \log_{10} (99.52 \times 10^{12}) \]
\[ L = 10 \left( \log_{10} (99.52) + 12 \right) \]

Calculating \( \log_{10} (99.52) \):
\[ \log_{10} (99.52) \approx 2 \]
Thus:
\[ L \approx 10 \left( 2 + 12 \right) = 10 \times 14 = 140 \, \text{dB} \]

### Final Answer:
The sound intensity level at a point 4.00 m away from the speaker is approximately **140 dB**.

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