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A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.

1. 12
2. 8
3. 6
4. 4

Correct answer: 6

Solution

The number of natural oscillations in a closed pipe can be determined using the formula for the fundamental frequency and its harmonics. For a closed pipe, the frequencies are given by the formula fₙ = n(v/4L), where n is an odd integer. Given the length of the pipe and the speed of sound, we can calculate the maximum harmonic that fits below 1250 Hz, which results in 6 possible oscillations.

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