Question

The distance between two consecutive points with phase difference of \(45^{\circ}\) in a wave of frequency \(300 \text{ Hz}\) is \(4.0 \text{ m}\). The velocity of the travelling wave is (in \( \text{km/s} \))

• A. \(1.6\)
• B. \(3.6\)
• C. \(4.8\)
• D. \(9.6\)

Hint:

A phase difference of $360^\circ$ ($2\pi$ radians) corresponds to a distance of one full wavelength ($\lambda$).

Answer 1

The Correct Option is D

Step 1: Phase and Path Difference
$\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.
Convert phase to radians: $45^\circ = \pi/4$.

Step 2: Find Wavelength
$\pi/4 = \frac{2\pi}{\lambda} \cdot 4 \implies \frac{1}{4} = \frac{8}{\lambda} \implies \lambda = 32 \text{ m}$.

Step 3: Find Velocity
$v = f\lambda = 300 \times 32 = 9600 \text{ m/s}$.
$v = 9.6 \text{ km/s}$.
Final Answer: (D)