A source emitting sound of frequency 100 Hz . It is placed in front of a wall at a distance of 2m from it. Also, if a detector is also placed in front of the wall at the same distance from it, then the minimum distance between the source and the detector for which detector detects a maximum of sound is
________ `m ( v _("sound") = 360 m//s)`

Explanation

To solve the problem step by step, we will analyze the situation involving the sound source, the wall, and the detector.

### Step 1: Understand the Setup
We have a sound source emitting sound at a frequency of 100 Hz, placed 2 m away from a wall. A detector is also placed 2 m away from the wall. We need to find the minimum distance \( x \) between the source and the detector for which the detector detects a maximum of sound.

### Step 2: Calculate the Wavelength
The speed of sound \( v \) is given as 360 m/s. The wavelength \( \lambda \) can be calculated using the formula:
\[ \lambda = \frac{v}{f} \]
where \( f \) is the frequency of the sound. Substituting the values:
\[ \lambda = \frac{360 \, \text{m/s}}{100 \, \text{Hz}} = 3.6 \, \text{m} \]

### Step 3: Determine Path Difference for Maximum Sound
For the detector to detect a maximum sound, the path difference \( \Delta x \) between the direct sound path and the sound reflected off the wall must be an integer multiple of the wavelength:
\[ \Delta x = n\lambda \quad (n = 0, 1, 2, \ldots) \]
For the first maximum, we take \( n = 1 \):
\[ \Delta x = \lambda = 3.6 \, \text{m} \]

### Step 4: Analyze the Path Lengths
Let \( x \) be the distance between the source and the detector. The total distance traveled by the sound from the source to the wall and then to the detector is:
\[ \text{Distance via wall} = 2 + 2 + x = 4 + x \]
The direct distance from the source to the detector is simply \( x \).

### Step 5: Set Up the Equation for Path Difference
The path difference is given by:
\[ \Delta x = (4 + x) - x = 4 \, \text{m} \]
We need this path difference to equal \( 3.6 \, \text{m} \):
\[ 4 = n \cdot 3.6 \]
Solving for \( n \):
\[ n = \frac{4}{3.6} \approx 1.11 \]
Since \( n \) must be an integer, we can check for \( n = 1 \) and \( n = 2 \).

### Step 6: Solve for Minimum Distance
Using \( n = 1 \):
\[ \Delta x = 3.6 \, \text{m} \]
Setting up the equation:
\[ 4 + x - x = 3.6 \]
This simplifies to:
\[ 4 = 3.6 + x \]
Thus:
\[ x = 4 - 3.6 = 0.4 \, \text{m} \]

### Step 7: Conclusion
The minimum distance between the source and the detector for which the detector detects a maximum of sound is:
\[ \boxed{0.4 \, \text{m}} \]