Step 1: Understanding the Question:
A closed organ pipe produces only odd harmonics.
We need to find how many overtones have frequencies less than or equal to the human hearing limit of \(19.5 \text{ kHz}\).
Step 2: Key Formula or Approach:
The frequency of the \(n^{th}\) harmonic for a closed pipe is given by:
\[ f_{k} = (2n - 1)f_{1} \]
where \(f_{1}\) is the fundamental frequency and \(n = 1, 2, 3, \dots\)
The \(p^{th}\) overtone corresponds to the \((2p + 1)^{th}\) harmonic.
Step 3: Detailed Explanation:
Given fundamental frequency \(f_{1} = 1500 \text{ Hz}\).
The hearing limit is \(f_{max} = 19.5 \text{ kHz} = 19500 \text{ Hz}\).
We set the condition for the highest audible harmonic:
\[ (2n - 1) \times 1500 \le 19500 \]
\[ 2n - 1 \le \frac{19500}{1500} \]
\[ 2n - 1 \le 13 \]
\[ 2n \le 14 \implies n \le 7 \]
The audible harmonics are for \(n = 1, 2, 3, 4, 5, 6, 7\).
These correspond to the fundamental (\(n=1\)) and the subsequent overtones.
Overtones start from \(n = 2\). So, the audible overtones are for \(n = 2, 3, 4, 5, 6, 7\).
The total number of audible overtones is \(7 - 1 = 6\).
The frequencies are:
1st Overtone (\(3 \times 1500 = 4500 \text{ Hz}\))
2nd Overtone (\(5 \times 1500 = 7500 \text{ Hz}\))
3rd Overtone (\(7 \times 1500 = 10500 \text{ Hz}\))
4th Overtone (\(9 \times 1500 = 13500 \text{ Hz}\))
5th Overtone (\(11 \times 1500 = 16500 \text{ Hz}\))
6th Overtone (\(13 \times 1500 = 19500 \text{ Hz}\))
Step 4: Final Answer:
The maximum number of overtones is \(6\).