Question

A stationary sound wave has a frequency of $165\ \text{Hz}$. If the speed of sound in air is $330\ \text{m/s}$, then the distance between a node and the adjacent antinode is

• A. $80\ \text{cm}$
• B. $50\ \text{cm}$
• C. $2\ \text{cm}$
• D. $20\ \text{cm}$

Hint:

In stationary waves, distance between a node and adjacent antinode is always one-fourth of the wavelength.

Answer 1

The Correct Option is B

Step 1: Find wavelength of the sound wave.
\[ \lambda = \dfrac{v}{f} = \dfrac{330}{165} = 2\ \text{m} \] Step 2: Relation between node and antinode.
The distance between a node and the adjacent antinode is $\dfrac{\lambda}{4}$.
Step 3: Calculate required distance.
\[ \dfrac{\lambda}{4} = \dfrac{2}{4} = 0.5\ \text{m} = 50\ \text{cm} \]