Question

Two points on a travelling wave having frequency 500 Hz and velocity \(300 \mathrm{~m} / \mathrm{s}\) are \(30^{\circ}\) out of phase, then the minimum distance between the two points is

• A. 0.02 m
• B. 0.01 m
• C. 0.05 m
• D. 0.1 m

Hint:

Find the wavelength from \(\lambda=v/f\), then use path difference \(=\dfrac{\lambda}{2\pi}\times\) phase difference.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Phase difference \(\phi\) and path difference \(\Delta x\) are related by \(\Delta x=\dfrac{\lambda}{2\pi}\phi\).

Step 2: Wavelength
\(\lambda=\dfrac{v}{f}=\dfrac{300}{500}=0.6\ \mathrm{m}\).

Step 3: Phase in radians
\(30^{\circ}=\dfrac{\pi}{6}\).

Step 4: Distance
\(\Delta x=\dfrac{0.6}{2\pi}\times\dfrac{\pi}{6}=0.05\ \mathrm{m}\).

Final Answer:
The minimum distance between the two points is 0.05 m. \[ \boxed{\text{(C) }0.05\text{ m}} \]