An open pipe is in resonance in its 2^{nd} ha | vedclass

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(C) For an open pipe,the frequency of the $2^{nd}$ harmonic is given by $f_1 = \frac{2v}{2L} = \frac{v}{L}$.When one end is closed,the pipe becomes a

Vedclass · Accepted Answer

$n = 5, f_2 = \frac{5}{4}f_1$

Vedclass · Answer

(C) For an open pipe,the frequency of the $2^{nd}$ harmonic is given by $f_1 = \frac{2v}{2L} = \frac{v}{L}$.When one end is closed,the pipe becomes a closed organ pipe. The resonant frequencies for a closed pipe are given by $f_n = \frac{nv}{4L}$,where $n$ is an odd integer $(n = 1, 3, 5, \dots)$.We are given that $f_2 > f_1$. Substituting $v = f_1 L$ into the closed pipe formula,we get $f_2 = \frac{n(f_1 L)}{4L} = \frac{n}{4}f_1$.Since $f_2 > f_1$,we must have $\frac{n}{4} > 1$,which implies $n > 4$.The smallest odd integer greater than $4$ is $n = 5$.Therefore,$f_2 = \frac{5}{4}f_1$ and $n = 5$.

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