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:

Key Exam Tip:
For standing waves, the distance between successive nodes (or antinodes) is $\lambda/2$. Use $v = f \lambda$ to calculate the wave velocity.

Answer 1

The Correct Option is B

In a stationary (standing) wave, the distance between two successive nodes or two successive antinodes is equal to half of the wavelength. \[ \text{Distance between successive nodes} = \frac{\lambda}{2} \] Given: \[ \frac{\lambda}{2} = 34\ \text{cm} \] Converting into SI unit: \[ 34\ \text{cm} = 0.34\ \text{m} \] Therefore, \[ \frac{\lambda}{2} = 0.34 \] \[ \lambda = 2 \times 0.34 \] \[ \lambda = 0.68\ \text{m} \] The relation between velocity, frequency, and wavelength is: \[ v = f\lambda \] where:
• $v$ = velocity of sound
• $f$ = frequency
• $\lambda$ = wavelength Given: \[ f = 500\ \text{Hz} \] \[ \lambda = 0.68\ \text{m} \] Substituting the values: \[ v = 500 \times 0.68 \] \[ v = 340\ \text{m/s} \] Hence, the velocity of sound is: \[ \boxed{340\ \text{m/s}} \] Final Answer: \(\boxed{B}\)