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

Answer

To find the frequency of the sound wave given the velocity of sound in air and the path and phase differences, we can follow these steps:

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

### Step 2: Rearrange the formula to find the wavelength.
We can rearrange the formula to solve for the wavelength (λ):
\[ \lambda = \frac{2\pi \Delta x}{\Delta \phi} \]

### Step 3: Substitute the known values.
Given:
- Path difference, Δx = 40 cm = 0.4 m
- Phase difference, Δφ = 1.6π

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

### Step 4: Simplify the equation.
The π cancels out:
\[ \lambda = \frac{2 \times 0.4}{1.6} = \frac{0.8}{1.6} = 0.5 \text{ m} \]

### Step 5: Use the wavelength to find the frequency.
We know the velocity of sound (v) is given as 330 m/s. The relationship between velocity (v), frequency (ν), and wavelength (λ) is:
\[ v = \lambda \nu \]
Rearranging to find frequency:
\[ \nu = \frac{v}{\lambda} \]

### Step 6: Substitute the known values for velocity and wavelength.
Substituting the values:
\[ \nu = \frac{330 \text{ m/s}}{0.5 \text{ m}} = 660 \text{ Hz} \]

### Conclusion:
The frequency of the wave is 660 Hz.

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