Question

If sound waves of frequency 169 Hz are incident horizontally on a perfectly rigid vertical wall, then the shortest distance from the wall at which the air particle will have maximum amplitude of vibration is:

• A. \(50\ cm\)
• B. \(25\ cm\)
• C. \(100\ cm\)
• D. \(75\ cm\)

Hint:

At a rigid wall: \[ \text{Node at wall} \] and the first antinode is located at \[ \lambda/4. \]

Answer 1

The Correct Option is A

Concept: Reflection from a rigid wall produces a stationary wave. At a rigid wall: \[ \text{Displacement node} \] is formed. The nearest displacement antinode occurs at \[ \frac{\lambda}{4}. \]

Step 1: Find wavelength.
Taking velocity of sound \[ v=338\,m/s. \] \[ \lambda = \frac{v}{f} = \frac{338}{169} = 2m. \]

Step 2: Locate first antinode.
\[ x=\frac{\lambda}{4} = \frac{2}{4} = 0.5m. \] \[ =50cm. \] Hence \[ \boxed{50cm}. \]