Question

A progressive wave of frequency $400 \text{ Hz}$ is travelling with velocity $336 \text{ m/s}$. How far apart are the two points on a wave which are $60^\circ$ out of phase?

• A. 0.12 m
• B. 0.14 m
• C. 0.21 m
• D. 0.28 m

Hint:

$360^\circ$ corresponds to one full wavelength ($\lambda$). $60^\circ$ is $1/6$th of a wavelength.

Answer 1

The Correct Option is B

Step 1: Concept
Phase difference $\phi$ and path difference $\Delta x$ are related by $\phi = \frac{2\pi}{\lambda} \Delta x$.

Step 2: Meaning
Wavelength $\lambda = v/f = 336 / 400 = 0.84 \text{ m}$. Phase $\phi = 60^\circ = \pi/3$ radians.

Step 3: Analysis
$\pi/3 = (2\pi / 0.84) \Delta x \implies 1/3 = (2 / 0.84) \Delta x \implies 0.84 / 6 = \Delta x$.

Step 4: Conclusion
$\Delta x = 0.14 \text{ m}$. Final Answer: (B)