Question

The path difference between two particles of a sound wave is \( 50 \) cm and the phase difference between them is \( 1.8\pi \). If the speed of sound in air is \( 340 \) m/s, the frequency of the sound wave is?

• A. \( 672 \) Hz
• B. \( 306 \) Hz
• C. \( 612 \) Hz
• D. \( 340 \) Hz

Hint:

To relate phase difference and path difference, use \( \Delta \phi = \frac{2\pi}{\lambda} \Delta x \). Then, apply \( f = v/\lambda \) to determine the frequency.

Answer 1

The Correct Option is C

Step 1: Relationship between path difference and phase difference 
The phase difference \( \Delta \phi \) and path difference \( \Delta x \) of a wave are related by: \[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \] Given: \[ \Delta \phi = 1.8\pi, \quad \Delta x = 50 \text{ cm} = 0.50 \text{ m} \] Step 2: Solve for wavelength \( \lambda \) 
Rearranging the equation: \[ \lambda = \frac{2\pi \times 0.50}{1.8\pi} \] \[ \lambda = \frac{1.0}{1.8} = 0.555 \text{ m} \] Step 3: Calculate frequency 
The frequency of a wave is given by: \[ f = \frac{v}{\lambda} \] where \( v = 340 \) m/s is the speed of sound in air. \[ f = \frac{340}{0.555} \] \[ f \approx 612 \text{ Hz} \] Thus, the frequency of the sound wave is \( 612 \) Hz.