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: Understanding the Question:
Phase difference \(\Delta \phi\) and path difference \(\Delta x\) are related via wavelength. We need to find the wavelength first to calculate wave velocity.
Step 2: Key Formula or Approach:
1) Relation: \(\Delta \phi = \frac{2\pi}{\lambda} \Delta x\).
2) Wave velocity: \(v = f \lambda\).
Step 3: Detailed Explanation:
Given:
Phase difference \(\Delta \phi = 45^{\circ} = \frac{\pi}{4} \text{ radians}\).
Path difference \(\Delta x = 4.0 \text{ m}\).
Frequency \(f = 300 \text{ Hz}\).
Using the phase difference relation:
\[ \frac{\pi}{4} = \frac{2\pi}{\lambda} \times 4.0 \]
Cancel \(\pi\) and solve for \(\lambda\):
\[ \frac{1}{4} = \frac{8.0}{\lambda} \implies \lambda = 32 \text{ m} \]
Now, calculate the velocity:
\[ v = f \times \lambda = 300 \times 32 = 9600 \text{ m/s} \]
Convert to \(\text{km/s}\):
\[ v = \frac{9600}{1000} \text{ km/s} = 9.6 \text{ km/s} \]
Step 4: Final Answer:
The velocity of the wave is \(9.6 \text{ km/s}\).