Question

Velocity of sound waves in air is 330 m/s. For a particular sound wave in air, a path difference of 40 cm is equivalent to phase difference of $1.6\, \pi$. The frequency of this wave

• A. 165 Hz
• B. 150 Hz
• C. 660 Hz
• D. 330 Hz

Answer 1

The Correct Option is C

To determine the frequency of the sound wave, we should first establish the relationship between path difference and phase difference for a wave. The formula for phase difference (\Delta \phi) in terms of path difference (\Delta x) is given by:

\Delta \phi = \frac{2\pi}{\lambda} \times \Delta x

where \lambda is the wavelength of the wave.

From the problem, we know that a path difference of 40 cm (0.4 m) corresponds to a phase difference of 1.6\pi. Therefore, we can write:

1.6\pi = \frac{2\pi}{\lambda} \times 0.4

Solving for \lambda, we get:

1.6 = \frac{2 \times 0.4}{\lambda}

\lambda = \frac{0.8}{1.6} = 0.5\ \text{m}

Now that we have the wavelength \lambda = 0.5\ \text{m}, we can find the frequency f using the wave equation:

v = f \lambda

Given the velocity of sound v = 330\ \text{m/s}, we can solve for f:

330 = f \times 0.5

f = \frac{330}{0.5} = 660\ \text{Hz}

Therefore, the frequency of the wave is 660 Hz.

Let's reason why the other options are incorrect:

• 165 Hz: This would correspond to a much longer wavelength, which is not consistent with the given path and phase difference.
• 150 Hz: Similarly, this frequency implies a larger wavelength not matching the phase difference condition.
• 330 Hz: While it is half of the correct answer, it doesn't satisfy the wave equation with the provided velocity and calculated wavelength.

In conclusion, based on the given information and calculations, the correct answer is 660 Hz.