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 step by step, we need to understand the conditions under which silence occurs due to destructive interference of sound waves. Here’s how we can approach the problem:

### Step 1: Understand the Condition for Silence
Silence occurs at a point when the two sound waves interfere destructively. This happens when the path difference between the two waves is an odd multiple of half the wavelength (λ/2).

### Step 2: Identify the Path Differences
The given path differences where silence occurs are:
- Path difference 1 (Δx1) = 12 cm = 0.12 m
- Path difference 2 (Δx2) = 36 cm = 0.36 m

### Step 3: Use the Formula for Path Difference
The path difference for destructive interference can be expressed as:
\[ \Delta x = (n + \frac{1}{2}) \lambda \]
where n is an integer (0, 1, 2, ...).

### Step 4: Calculate the Wavelength
From the path differences, we can set up the equations:
1. For Δx1 = 0.12 m:
\[ 0.12 = (n_1 + \frac{1}{2}) \lambda \]
2. For Δx2 = 0.36 m:
\[ 0.36 = (n_2 + \frac{1}{2}) \lambda \]

### Step 5: Solve for Wavelength
To find the wavelength, we can rearrange the equations:
From the first equation:
\[ \lambda = \frac{0.12}{n_1 + \frac{1}{2}} \]
From the second equation:
\[ \lambda = \frac{0.36}{n_2 + \frac{1}{2}} \]

Setting these equal gives:
\[ \frac{0.12}{n_1 + \frac{1}{2}} = \frac{0.36}{n_2 + \frac{1}{2}} \]

Cross-multiplying and simplifying will help us find a relationship between \( n_1 \) and \( n_2 \).

### Step 6: Calculate the Speed of Sound and Frequency
The speed of sound (v) is given as 330 m/s. The relationship between speed, frequency (f), and wavelength (λ) is:
\[ v = f \lambda \]
Thus, we can express frequency as:
\[ f = \frac{v}{\lambda} \]

### Step 7: Find the Maximum Frequency
To find the maximum possible frequency, we need to use the minimum wavelength that satisfies the conditions for both path differences.

Assuming the smallest value of n (n = 0) for maximum frequency:
\[ \lambda = 0.12 \text{ m} \text{ (for n = 0)} \]
Then:
\[ f = \frac{330}{0.12} = 2750 \text{ Hz} \]

### Step 8: Verify with Other Values of n
We can check for n = 1 and n = 2 to ensure that we have the maximum frequency. However, since the path differences are multiples of 12 cm, we can conclude that the maximum frequency occurs at the minimum wavelength.

### Final Answer
The maximum possible frequency of the source is **2750 Hz**.

---