Question

Two progressive waves are travelling towards each other with velocity \(50 \, \text{m s}^{-1}\) and frequency \(200 \, \text{Hz}\). The distance between two consecutive antinodes is

• A. \(0.031 \, \text{m}\)
• B. \(0.125 \, \text{m}\)
• C. \(0.250 \, \text{m}\)
• D. \(0.0625 \, \text{m}\)

Hint:

In stationary waves, the separation between successive antinodes (or nodes) is always half the wavelength.

Answer 1

The Correct Option is B

Step 1: Calculating wavelength of the waves.
Wave velocity is given by \[ v = f\lambda. \] Substituting given values, \[ \lambda = \frac{v}{f} = \frac{50}{200} = 0.25 \, \text{m}. \]
Step 2: Distance between consecutive antinodes.
When two identical waves travel in opposite directions, a stationary wave is formed. The distance between two consecutive antinodes is \[ \frac{\lambda}{2}. \]
Step 3: Final calculation.
\[ \text{Distance} = \frac{0.25}{2} = 0.125 \, \text{m}. \]
Step 4: Conclusion.
The distance between two consecutive antinodes is \(0.125 \, \text{m}\).