A closed organ pipe has a fundamental frequen | vedclass

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(A) For a closed organ pipe,the resonant frequencies are odd multiples of the fundamental frequency,given by $f_n = (2n + 1)f_0$,where $n = 0, 1, 2, \

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$6$

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(A) For a closed organ pipe,the resonant frequencies are odd multiples of the fundamental frequency,given by $f_n = (2n + 1)f_0$,where $n = 0, 1, 2, \dots$ and $f_0 = 1500\, Hz$.We need to find the number of overtones such that $f_n \leq 20,000\, Hz$.$(2n + 1) 	imes 1500 \leq 20,000$$2n + 1 \leq \frac{20,000}{1500} = 13.33$$2n \leq 12.33 \implies n \leq 6.16$.Since $n$ must be an integer,the possible values for $n$ are $0, 1, 2, 3, 4, 5, 6$.Here,$n=0$ corresponds to the fundamental frequency.The overtones correspond to $n = 1, 2, 3, 4, 5, 6$.Thus,there are $6$ overtones that can be heard.

$A$ closed organ pipe has a fundamental frequency of $1.5\, kHz$. The number of overtones that can be distinctly heard by a person with this organ pipe will be: (Assume that the highest frequency a person can hear is $20,000\, Hz$)

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