Question

Sound waves of frequency $600\ \text{Hz}$ fall normally on a perfectly reflecting wall. The shortest distance from the wall at which all particles will have maximum amplitude of vibration is (speed of sound $=300\ \text{ms}^{-1}$)}

• A. $\dfrac{1}{4}\ \text{m}$
• B. $\dfrac{1}{8}\ \text{m}$
• C. $\dfrac{3}{8}\ \text{m}$
• D. $\dfrac{7}{8}\ \text{m}$

Hint:

For sound reflection at a rigid wall, the wall is a displacement node. The nearest point of maximum particle vibration lies at a distance $\lambda/4$ from the wall.

Answer 1

The Correct Option is B

Step 1: Find wavelength of sound wave. Relation between wave speed, frequency and wavelength: \[ v=f\lambda \] Given: \[ v=300\ \text{ms}^{-1}, \qquad f=600\ \text{Hz} \] So, \[ \lambda=\frac{v}{f}=\frac{300}{600}=0.5\ \text{m} \]
Step 2: Understand reflection from rigid wall. When sound reflects from a perfectly reflecting rigid wall, a stationary wave is formed. For displacement of air particles:

• Wall acts as a displacement node
• Maximum amplitude occurs at displacement antinode

Step 3: Find nearest antinode from wall. Distance between a node and nearest antinode is: \[ \frac{\lambda}{4} \] Hence shortest distance from wall for maximum vibration amplitude: \[ x=\frac{0.5}{4}=0.125\ \text{m} \] \[ x=\frac{1}{8}\ \text{m} \]
Step 4: Conclude the correct option. Therefore, the required shortest distance is: \[ \boxed{\frac{1}{8}\ \text{m \]