Question

Two points on a travelling wave having frequency \(500\) Hz and velocity \(300\) m/s are \(60^{\circ }\) out of phase, then the minimum distance between the two points is

• A. \(0.2\) m
• B. \(0.1\) m
• C. \(0.5\) m
• D. \(0.4\) m

Hint:

Find the wavelength, then use path difference equal to phase difference times wavelength over 2 pi.

Answer 1

The Correct Option is B

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

Step 2: Find the Wavelength
\(\lambda=\frac{v}{f}=\frac{300}{500}=0.6\) m.

Step 3: Convert the Phase
\(\Delta\phi=60^{\circ}=\frac{\pi}{3}\) rad.

Step 4: Calculation
\(\Delta x=\frac{0.6}{2\pi}\times\frac{\pi}{3}=0.1\) m, option (B).

Final Answer:
The minimum distance between the two points is 0.1 m. \[ \boxed{\text{(B) }0.1\ \text{m}} \]