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Step-by-Step Solution
Step 1: Identify the Type of Organ Pipe
The pipe is closed at one end and open at the other. Such a system is known as a “closed organ pipe.”
Step 2: Write Down the Relevant Formula
The fundamental frequencies (natural frequencies) for a closed organ pipe of length $L$ are given by:
$f_n = \frac{(2n - 1)\,v}{4L},$
where
$n$ is the harmonic number (an integer starting from 1),
$v$ is the speed of sound in air,
$L$ is the length of the pipe.
Step 3: Plug in Known Values
From the problem:
$L = 0.85\,\text{m}$,
$v = 340\,\text{m/s}$,
Maximum permissible frequency $f_{max} = 1250\,\text{Hz}$.
We set up the inequality for $f_n$ to be less than or equal to $1250\,\text{Hz}$:
$\frac{(2n - 1)\,340}{4 \times 0.85} \le 1250.$
Step 4: Simplify the Inequality
$\frac{(2n - 1)\,340}{3.4} \le 1250
\quad\Longrightarrow\quad
(2n - 1) \le \frac{1250 \times 3.4}{340}.
$
Now, $4 \times 0.85 = 3.4,$ and $1250 \times 3.4 = 4250,$ so:
$(2n - 1) \le \frac{4250}{340} = 12.5.
Step 5: Solve for the Harmonic Number n
$(2n - 1) \le 12.5
\quad\Longrightarrow\quad
2n \le 13.5
\quad\Longrightarrow\quad
n \le 6.75.
$
Since $n$ must be a positive integer, the possible values are $n = 1, 2, 3, 4, 5, 6.$
Step 6: Count the Possible Frequencies
There are 6 possible natural oscillations of the air column in the pipe that lie below $1250\,\text{Hz}$.
Answer
6
