Two sound waves have intensities of 10 and `500mu//cm^(2)`. How many desibel is the second sound louder than the first?

Options

A. 7dB
B. 1.7dB
C. 2.7dB
D. 3.7dB

Answer

To find out how many decibels the second sound is louder than the first, we can use the formula for sound intensity level in decibels (dB):

1. **Identify the intensities of the two sound waves:**
- Let \( I_1 = 10 \, \mu \text{W/cm}^2 \)
- Let \( I_2 = 500 \, \mu \text{W/cm}^2 \)

2. **Use the formula for sound intensity level:**
The sound level \( L \) in decibels is given by:
\[ L = 10 \log \left( \frac{I}{I_0} \right) \]
where \( I_0 \) is the reference intensity (usually \( I_0 = 10^{-12} \, \text{W/m}^2 \)).

3. **Calculate the sound levels \( L_1 \) and \( L_2 \):**
- For the first sound:
\[ L_1 = 10 \log \left( \frac{I_1}{I_0} \right) \]
- For the second sound:
\[ L_2 = 10 \log \left( \frac{I_2}{I_0} \right) \]

4. **Find the difference in sound levels:**
The difference in sound levels is given by:
\[ L_2 - L_1 = 10 \log \left( \frac{I_2}{I_1} \right) \]

5. **Substitute the values of \( I_1 \) and \( I_2 \):**
\[ L_2 - L_1 = 10 \log \left( \frac{500}{10} \right) = 10 \log(50) \]

6. **Simplify \( \log(50) \):**
We can express \( \log(50) \) as:
\[ \log(50) = \log(10 \times 5) = \log(10) + \log(5) = 1 + \log(5) \]
The approximate value of \( \log(5) \) is about \( 0.7 \).

7. **Calculate the difference in decibels:**
\[ L_2 - L_1 = 10 \left( 1 + 0.7 \right) = 10 \times 1.7 = 17 \, \text{dB} \]

Thus, the second sound is **17 dB** louder than the first.