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EEJ MAIN Physics QUESTION #7029
Question 1
A pipe with a length of $85\text{ cm}$ is closed at one end. Determine the total number of natural oscillations possible for the air column in this pipe that have frequencies below $1250\text{ Hz}$. (Take the speed of sound as $340\text{ m/s}$).
  • $6$
  • $4$✔️
  • $12$
  • $8$
Correct Answer Explanation
For a closed pipe, the resonance frequencies are $f_n = \frac{(2n-1)v}{4L}$. Here $L = 0.85\text{ m}$ and $v = 340\text{ m/s}$. The fundamental frequency $f_1 = \frac{340}{4 \times 0.85} = 100\text{ Hz}$. The subsequent odd harmonics are $300\text{ Hz}$, $500\text{ Hz}$, $700\text{ Hz}$, $900\text{ Hz}$, and $1100\text{ Hz}$. The next is $1300\text{ Hz}$, which exceeds the limit. Thus, there are $6$ possible oscillations (including fundamental).