A dog while barking delivers about 1 mW of power. If this power is uniformly distributed over a hemispherical area, what is the sound level at a distance of 5 m? What would the sound level be if instead of 1 dog, 5 dogs start barking at the same time each delivering 1 mW of power

Options

A. 68 dB, 75 dB
B. 58 dB, 68 dB
C. 48 dB, 58 dB
D. 38 dB, 28 dB

Answer

To solve the problem step by step, we will first calculate the sound level produced by one dog barking and then determine the sound level when five dogs bark simultaneously.

### Step 1: Calculate the intensity of sound produced by one dog.
The power delivered by one dog is given as:
\[ P = 1 \text{ mW} = 1 \times 10^{-3} \text{ W} \]

The sound is uniformly distributed over a hemispherical area. The formula for the surface area \( A \) of a hemisphere is:
\[ A = 2\pi r^2 \]
where \( r \) is the distance from the source (5 m in this case).

Substituting the value of \( r \):
\[ A = 2\pi (5)^2 = 2\pi \times 25 = 50\pi \text{ m}^2 \]

Now, substituting the value of \( \pi \approx 3.14 \):
\[ A \approx 50 \times 3.14 = 157 \text{ m}^2 \]

The intensity \( I \) is given by the formula:
\[ I = \frac{P}{A} \]
Substituting the values:
\[ I = \frac{1 \times 10^{-3}}{157} \approx 6.36 \times 10^{-6} \text{ W/m}^2 \]

### Step 2: Calculate the sound level in decibels.
The reference intensity \( I_0 \) is:
\[ I_0 = 10^{-12} \text{ W/m}^2 \]

The sound level \( L \) in decibels is calculated using the formula:
\[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \]

Substituting the values:
\[ L = 10 \log_{10} \left( \frac{6.36 \times 10^{-6}}{10^{-12}} \right) \]
\[ L = 10 \log_{10} (6.36 \times 10^6) \]
Using the property of logarithms:
\[ L = 10 \left( \log_{10} (6.36) + \log_{10} (10^6) \right) \]
\[ L = 10 \left( \log_{10} (6.36) + 6 \right) \]
Using a calculator, \( \log_{10} (6.36) \approx 0.803 \):
\[ L \approx 10 \left( 0.803 + 6 \right) \]
\[ L \approx 10 \times 6.803 \]
\[ L \approx 68.03 \text{ dB} \]

### Step 3: Calculate the sound level when five dogs bark simultaneously.
When five dogs bark, the total power \( P_2 \) is:
\[ P_2 = 5 \times 1 \text{ mW} = 5 \times 10^{-3} \text{ W} \]

The intensity \( I_2 \) for five dogs is:
\[ I_2 = \frac{P_2}{A} = \frac{5 \times 10^{-3}}{157} \approx 3.18 \times 10^{-5} \text{ W/m}^2 \]

Now, we calculate the new sound level \( L_2 \):
\[ L_2 = 10 \log_{10} \left( \frac{I_2}{I_0} \right) \]
Using the ratio of intensities:
\[ L_2 = L_1 + 10 \log_{10} (5) \]
Where \( L_1 \approx 68 \text{ dB} \) from the previous calculation.

Calculating \( 10 \log_{10} (5) \):
\[ \log_{10} (5) \approx 0.699 \]
So,
\[ 10 \log_{10} (5) \approx 6.99 \]

Thus,
\[ L_2 \approx 68 + 7 \approx 75 \text{ dB} \]

### Final Answer:
- The sound level at a distance of 5 m from one dog barking is approximately **68 dB**.
- The sound level when five dogs bark simultaneously is approximately **75 dB**.