Two sound waves emerging from a source reach a point simultaneously along two paths. When the path difference is `12 cm ` or `36 cm`, then there is a silence at that point. If the speed of sound in air be `330 m//s`, then calculate maximum possible frequency of the source.

Explanation

To solve the problem, we need to find the maximum possible frequency of the sound source given the conditions of path difference and the speed of sound. Let's break down the solution step by step.

### Step 1: Understand the condition for silence
When two sound waves interfere destructively, they create silence at a point. This occurs when the path difference between the two waves is an odd multiple of half wavelengths. Mathematically, this can be expressed as:
\[ \Delta x = \left(n + \frac{1}{2}\right) \lambda \]
where \( n \) is an integer (0, 1, 2, ...).

### Step 2: Identify the path differences
Given path differences are:
- \( \Delta x_1 = 12 \, \text{cm} = 0.12 \, \text{m} \)
- \( \Delta x_2 = 36 \, \text{cm} = 0.36 \, \text{m} \)

### Step 3: Set up the equations
For both path differences, we can set up the equations:
1. For \( \Delta x_1 = 0.12 \, \text{m} \):
\[ 0.12 = \left(n_1 + \frac{1}{2}\right) \lambda \]
2. For \( \Delta x_2 = 0.36 \, \text{m} \):
\[ 0.36 = \left(n_2 + \frac{1}{2}\right) \lambda \]

### Step 4: Find the wavelength
From the two equations, we can express the wavelength \( \lambda \) in terms of \( n_1 \) and \( n_2 \):
\[ \lambda = \frac{0.12}{n_1 + \frac{1}{2}} = \frac{0.36}{n_2 + \frac{1}{2}} \]

### Step 5: Solve for \( n_1 \) and \( n_2 \)
We can rearrange the equations to find a relationship between \( n_1 \) and \( n_2 \):
\[ 0.12(n_2 + \frac{1}{2}) = 0.36(n_1 + \frac{1}{2}) \]
Expanding and simplifying gives:
\[ 0.12n_2 + 0.06 = 0.36n_1 + 0.18 \]
\[ 0.12n_2 - 0.36n_1 = 0.12 \]
\[ n_2 = 3n_1 + 1 \]

### Step 6: Substitute values
To find the maximum frequency, we need to find the maximum wavelength. The maximum occurs when \( n_1 \) is minimized (i.e., \( n_1 = 0 \)):
\[ n_2 = 1 \]
Substituting \( n_1 = 0 \) into the equation for \( \lambda \):
\[ \lambda = 0.12 \, \text{m} \]

### Step 7: Calculate frequency
Using the speed of sound \( v = 330 \, \text{m/s} \) and the wavelength \( \lambda \):
\[ f = \frac{v}{\lambda} = \frac{330}{0.12} \]
Calculating gives:
\[ f = 2750 \, \text{Hz} \]

### Final Answer
The maximum possible frequency of the source is:
\[ \boxed{2750 \, \text{Hz}} \]