An organ pipe closed at one end has fundamental frequency of 1500 Hz. The maximum number of overtones generated by this pipe a normal person can hear is (neglect end connection)

Question

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The fundamental frequency of the closed organ pipe is $f_{0} = 1500$ Hz.

For a closed organ pipe, only odd harmonics are present. The frequency of the $n$ -th overtone is given by:

$f_{n} = (2 n + 1) f_{0}$

The maximum audible frequency for a normal human ear is $20000$ Hz. Therefore, the frequency of the $n$ -th overtone must satisfy:

$(2 n + 1) f_{0} \leq 20000$

Substituting $f_{0} = 1500$ Hz:

$(2 n + 1) (1500) \leq 20000$

$2 n + 1 \leq \frac{20000}{1500}$

$2 n + 1 \leq 13.33$

$2 n \leq 12.33$

$n \leq 6.16$

Since the number of overtones $n$ must be an integer, the maximum possible value for $n$ is $6$ .

Thus, a normal person can hear a maximum of $6$ overtones.

Answer: $6$

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