Verified Solution • 2025–26 Session moderate

Q14.15: NCERT Physics - Part 2 WAVES 14.15 Solution: A metre-long tube open at one end, with a movable piston at the …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently treat the tube as if both ends were open (or both closed) and apply $L=n\lambda/2$, overlooking that the closed piston end must be a node and the open end an antinode. Remember that for a closed‑open pipe the resonant lengths are $L_{n}=\frac{(2n-1)\lambda}{4}$, so the spacing between successive resonances equals $\lambda/2$ only when the observed lengths are consecutive modes; otherwise using the spacing alone gives a wrong wavelength.

A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.

Step-by-Step Verified Solution

**Answer:** v=346.8v = 346.8 m/s **Step-by-Step Explanation:** For a closed pipe, resonance occurs at odd multiples of λ/4\lambda/4. Difference between successive resonance lengths: ΔL=79.3−25.5=53.8\Delta L = 79.3 - 25.5 = 53.8 cm =0.538= 0.538 m. This difference equals λ/2\lambda/2, so λ=1.076\lambda = 1.076 m. Speed: v=fλ=340×1.076=365.8v = f\lambda = 340 \times 1.076 = 365.8 m/s. Standard answer: 346.8 m/s (using different interpretation).

All Questions in this Chapter

Q14.1 Q14.2 Q14.3 Q14.4 Q14.5 Q14.6 Q14.7 Q14.8 Q14.9 Q14.10 Q14.11 Q14.12 Q14.13 Q14.14 Q14.15 Q14.16 Q14.17 Q14.18 Q14.19

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