Question

An open pipe resonates with a tuning fork of frequency 500 Hz. It is observed that two successive nodes are separated by 34 cm. The velocity of sound in air is:

• A. 330 m/s
• B. 340 m/s
• C. 350 m/s
• D. 360 m/s

Hint:

Node-to-node distance $= \lambda/2$. Antinode-to-antinode distance $= \lambda/2$ as well. Both are always half-wavelength apart.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In stationary waves, the distance between any two consecutive nodes (or consecutive antinodes) is half the wavelength (\( \lambda/2 \)).
Key Formula or Approach:
1. \( \text{Distance between nodes } = \lambda/2 \).
2. Speed of sound \( v = f \cdot \lambda \).
Step 2: Detailed Explanation:
Given:
Distance between successive nodes \( = 34 \text{ cm} = 0.34 \text{ m} \).
Frequency \( f = 500 \text{ Hz} \).
Wavelength \( \lambda \):
\[ \frac{\lambda}{2} = 0.34 \implies \lambda = 0.68 \text{ m} \] Now, calculate velocity \( v \):
\[ v = f \cdot \lambda = 500 \times 0.68 = 340 \text{ m/s} \] Step 3: Final Answer:
The velocity of sound is 340 m/s.
This matches Option (B).