Search any question & find its solution
Question:
Answered & Verified by Expert
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 $340 \mathrm{~Hz}$ ) when the tube length is $25.5 \mathrm{~cm}$ or $79.3 \mathrm{~cm}$. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
Solution:
1521 Upvotes
Verified Answer
[QUESTION]
One end of the tube is open and the other end is closed by the piston, so it behaves as a closed organ pipe, which produces only odd harmonics.
$\therefore \quad$ The pipe is in resonance with the fundamental mode and the 3rd harmonic $(\because 79.3 \approx 3 \times 25.5)$
For fundamental mode, $\frac{\lambda}{4}-l_1-25.5$ $\therefore \lambda=4 \times 25.5=102 \mathrm{~cm}=1.02 \mathrm{~m}$
Speed $=v=v \lambda=340 \times 1.02=346.8 \mathrm{~m} / \mathrm{s}$
One end of the tube is open and the other end is closed by the piston, so it behaves as a closed organ pipe, which produces only odd harmonics.
$\therefore \quad$ The pipe is in resonance with the fundamental mode and the 3rd harmonic $(\because 79.3 \approx 3 \times 25.5)$
For fundamental mode, $\frac{\lambda}{4}-l_1-25.5$ $\therefore \lambda=4 \times 25.5=102 \mathrm{~cm}=1.02 \mathrm{~m}$
Speed $=v=v \lambda=340 \times 1.02=346.8 \mathrm{~m} / \mathrm{s}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.