Question

A simple harmonic progressive wave is represented by \(y = Asin(120πt+3x)\). The distance between two points on the wave at a phase difference of \(\frac{π}{3}\) radian is

• A. \(\frac{π}{6}\) m
• B. \(\frac{π}{9}\) m
• C. \(\frac{2π}{9}\) m
• D. \(\frac{π}{18}\) m

Hint:

Phase difference = (2 pi / lambda) x path difference = k x path difference.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Comparing \(y = A\sin(120\pi t + 3x)\) with \(y = A\sin(\omega t + kx)\) gives wave number \(k = 3\) rad/m.

Step 2: Compute
The phase difference between two points at a distance \(\Delta x\) is \(\Delta\phi = k\,\Delta x\):
\[ \Delta x = \frac{\Delta\phi}{k} = \frac{\pi/3}{3} = \frac{\pi}{9} \text{ m} \]
Option (A) would come from using k = 2 instead of 3.

Final Answer:
The distance is \(\frac{\pi}{9}\) m, option (B). \[ \boxed{\frac{\pi}{9} \text{ m}} \]