Two identical sounds `s_(1) and s_(2)` reach at a point P phase. The resultant loudness at a point P is not higher than the loudness of `s_(1)`. The value of n is

Options

A. 2
B. 4
C. 5
D. 6

Answer

To solve the problem, we need to analyze the situation where two identical sound sources \( s_1 \) and \( s_2 \) reach a point \( P \) in phase, and the resultant loudness at point \( P \) is not higher than the loudness of \( s_1 \).

### Step-by-Step Solution:

1. **Understanding the Phase Condition**:
Since \( s_1 \) and \( s_2 \) are in phase, the phase difference \( \Delta \phi \) between them is 0. This means that their sound waves will constructively interfere at point \( P \).

2. **Resultant Intensity Calculation**:
The resultant intensity \( I_R \) when two identical sound sources are in phase can be calculated using the formula:
\[ I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \]
Since both sources are identical, we can denote their intensity as \( I_0 \):
\[ I_R = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0} = 2I_0 + 2I_0 = 4I_0 \]

3. **Loudness Calculation**:
The loudness \( L \) of a sound is related to its intensity \( I \) by the formula:
\[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \]
Here, \( I_0 \) is the reference intensity. For our resultant intensity:
\[ L_R = 10 \log_{10} \left( \frac{4I_0}{I_0} \right) = 10 \log_{10} (4) \]

4. **Calculating \( \log_{10} (4) \)**:
We know that:
\[ \log_{10} (4) = \log_{10} (2^2) = 2 \log_{10} (2) \]
The approximate value of \( \log_{10} (2) \) is about 0.301, thus:
\[ \log_{10} (4) \approx 2 \times 0.301 = 0.602 \]

5. **Final Loudness Calculation**:
Substituting back into the loudness equation:
\[ L_R = 10 \times 0.602 \approx 6.02 \]
Since the problem states that the resultant loudness at point \( P \) is not higher than the loudness of \( s_1 \), we can conclude that \( n \) must be equal to 6.

### Conclusion:
The value of \( n \) is 6.