The number of possible natural oscillations of the air column in a pipe closed at one end of a length of $85 \

Question

The number of possible natural oscillations of the air column in a pipe closed at one end of a length of $85 	ext{ cm}$ whose frequencies lie below $1250 	ext{ Hz}$ is: (velocity of sound $340 	ext{ ms}^{-1}$)

Accepted Answer

1. **Identify the formula for a closed organ pipe:** The natural frequencies for a pipe closed at one end are odd harmonics, given by $f = (2n - 1) \frac{v}{4L}$, where $v$ is the speed of sound, $L$ is the length of the pipe, and $n = 1, 2, 3, \dots$ .
2. **Calculate the fundamental frequency ($f_1$):**
   Given $v = 340 	ext{ m/s}$ and $L = 85 	ext{ cm} = 0.85 	ext{ m}$.
   $$f_1 = \frac{v}{4L} = \frac{340}{4 	imes 0.85} = \frac{340}{3.4} = 100 	ext{ Hz}$$
3. **Determine the possible frequencies:** A closed pipe produces only odd multiples of the fundamental frequency. The frequencies are:
   $f_1 = 1 	imes 100 = 100 	ext{ Hz}$
   $f_2 = 3 	imes 100 = 300 	ext{ Hz}$
   $f_3 = 5 	imes 100 = 500 	ext{ Hz}$
   $f_4 = 7 	imes 100 = 700 	ext{ Hz}$
   $f_5 = 9 	imes 100 = 900 	ext{ Hz}$
   $f_6 = 11 	imes 100 = 1100 	ext{ Hz}$
   $f_7 = 13 	imes 100 = 1300 	ext{ Hz}$
4. **Count frequencies below $1250 	ext{ Hz}$:** The possible natural frequencies below $1250 	ext{ Hz}$ are 100, 300, 500, 700, 900, and 1100 Hz. There are a total of 6 such frequencies.

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