The velocity of sound waves in air is `330m//s`. For a particluar sound in air, a path difference of `40 cm` is equivalent to a phase difference of `1.6pi`. The frequency of this wave is

Options

A. `165 hz`
B. `150 Hz`
C. `660 Hz`
D. `330 Hz`

Explanation

To find the frequency of the sound wave, we can follow these steps:

### Step 1: Understand the relationship between phase difference, path difference, and wavelength
The phase difference (\(\Delta \phi\)) is related to the path difference (\(\Delta x\)) and the wavelength (\(\lambda\)) by the formula:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]

### Step 2: Substitute the known values into the equation
Given:
- Phase difference, \(\Delta \phi = 1.6\pi\)
- Path difference, \(\Delta x = 40 \text{ cm} = 0.4 \text{ m}\)

Substituting these values into the equation:
\[ 1.6\pi = \frac{2\pi}{\lambda} \cdot 0.4 \]

### Step 3: Simplify the equation
We can cancel \(\pi\) from both sides:
\[ 1.6 = \frac{2 \cdot 0.4}{\lambda} \]

### Step 4: Solve for the wavelength (\(\lambda\))
Rearranging the equation to solve for \(\lambda\):
\[ \lambda = \frac{2 \cdot 0.4}{1.6} \]
Calculating this gives:
\[ \lambda = \frac{0.8}{1.6} = 0.5 \text{ m} \]

### Step 5: Use the wave velocity to find the frequency
The relationship between velocity (\(v\)), frequency (\(f\)), and wavelength (\(\lambda\)) is given by:
\[ v = f \cdot \lambda \]
We can rearrange this to find frequency:
\[ f = \frac{v}{\lambda} \]

### Step 6: Substitute the known values to find frequency
Given the velocity of sound in air, \(v = 330 \text{ m/s}\) and \(\lambda = 0.5 \text{ m}\):
\[ f = \frac{330}{0.5} = 660 \text{ Hz} \]

### Final Answer
The frequency of the wave is \(660 \text{ Hz}\).

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