In an organ pipe closed at one end, the sum of the frequencies of first three overtones is 3930 Hz . The frequency of the fundamental mode of organ pipe is

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For an organ pipe closed at one end, the frequencies of the overtones are odd multiples of the fundamental frequency $n$ .

First overtone $n_{1} = 3 n$
Second overtone $n_{2} = 5 n$
Third overtone $n_{3} = 7 n$

Given that the sum of the frequencies of the first three overtones is $3930 Hz$ .

$n_{1} + n_{2} + n_{3} = 3930$

$3 n + 5 n + 7 n = 3930$

$15 n = 3930$

$n = \frac{3930}{15} = 262 Hz$

Answer: 262 Hz

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