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(A) In a closed organ pipe of length $l$,the allowed frequencies are given by $
u_n = (2n + 1) \frac{v}{4l}$,where $n = 0, 1, 2, \dots$ represents th

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$2p + 1$

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(A) In a closed organ pipe of length $l$,the allowed frequencies are given by $
u_n = (2n + 1) \frac{v}{4l}$,where $n = 0, 1, 2, \dots$ represents the mode of vibration.For $n = 0$,$
u_0 = \frac{v}{4l}$ (Fundamental frequency or first harmonic).For $n = 1$,$
u_1 = 3 \frac{v}{4l}$ (First overtone or third harmonic).For $n = 2$,$
u_2 = 5 \frac{v}{4l}$ (Second overtone or fifth harmonic).In general,for the $p^{th}$ overtone,we set $n = p$. The frequency is $
u_p = (2p + 1) \frac{v}{4l}$.Since the fundamental frequency $
u_0 = \frac{v}{4l}$,we have $
u_p = (2p + 1) 
u_0$.Thus,the $p^{th}$ overtone corresponds to the $(2p + 1)^{th}$ harmonic.

In the case of a closed organ pipe,which harmonic is the $p^{th}$ overtone?

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