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:
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:
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Node-to-node distance $= \lambda/2$. Antinode-to-antinode distance $= \lambda/2$ as well. Both are always half-wavelength apart.
Updated On: May 29, 2026
- 330 m/s
- 340 m/s
- 350 m/s
- 360 m/s
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The Correct Option is B
Solution and Explanation
Step 1: Concept
In a standing wave, the distance between two successive nodes equals half the wavelength: $\dfrac{\lambda}{2}$.
Step 2: Meaning
Distance between nodes $= 34$ cm $= 0.34$ m and frequency $f = 500$ Hz.
Step 3: Analysis
Find the wavelength: \[\frac{\lambda}{2} = 0.34\ \text{m} \implies \lambda = 0.68\ \text{m}.\] Apply the wave equation $v = f\lambda$: \[v = 500 \times 0.68 = 340\ \text{m/s}.\]
Step 4: Conclusion
The velocity of sound in air is 340 m/s. Final Answer: (B)
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