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Step-by-Step Solution
Step 1: Identify the System
The question involves a closed organ pipe of length 85 cm, and we are asked to find how many possible natural frequencies are below 1250 Hz. In a pipe closed at one end, only odd harmonics of the fundamental frequency are produced.
Step 2: Write the Formula for the Fundamental Frequency
The fundamental frequency $f$ of a pipe closed at one end is given by:
$$f = \frac{v}{4L},$$
where
$v$ is the speed of sound in air.
$L$ is the length of the pipe.
Step 3: Substitute Known Values to Find the Fundamental Frequency
Given:
Speed of sound, $v = 340 \, \text{m/s}$
Length of the pipe, $L = 85 \, \text{cm} = 0.85 \, \text{m}$
Substitute these into the formula:
$$f = \frac{340}{4 \times 0.85} = 100 \, \text{Hz}.$$
Thus, the fundamental frequency is $100 \,\text{Hz}.$
Step 4: General Formula for Higher Frequencies in a Closed Pipe
The natural frequencies of a pipe closed at one end occur at odd multiples of the fundamental frequency. The $n$th frequency $f_n$ is given by:
$$f_n = (2n - 1) \, f,$$
where $n = 1, 2, 3, \dots$
Step 5: List the Frequencies Below 1250 Hz
Since $f = 100 \,\text{Hz},$ the odd harmonics are:
$$f_1 = (2 \times 1 - 1) \times 100 = 1 \times 100 = 100 \,\text{Hz}$$
$$f_2 = (2 \times 2 - 1) \times 100 = 3 \times 100 = 300 \,\text{Hz}$$
$$f_3 = (2 \times 3 - 1) \times 100 = 5 \times 100 = 500 \,\text{Hz}$$
$$f_4 = (2 \times 4 - 1) \times 100 = 7 \times 100 = 700 \,\text{Hz}$$
$$f_5 = (2 \times 5 - 1) \times 100 = 9 \times 100 = 900 \,\text{Hz}$$
$$f_6 = (2 \times 6 - 1) \times 100 = 11 \times 100 = 1100 \,\text{Hz}$$
$$f_7 = (2 \times 7 - 1) \times 100 = 13 \times 100 = 1300 \,\text{Hz}$$
Notice that $f_7 = 1300 \,\text{Hz}$ exceeds 1250 Hz, so it is not included.
Step 6: Count the Number of Frequencies Below 1250 Hz
The frequencies under 1250 Hz are 100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, and 1100 Hz. There are 6 such frequencies.
Step 7: Conclusion
Therefore, the number of possible natural oscillations of the air column in the pipe that lie below 1250 Hz is 6.
