The air column in a pipe closed at one end is made to vibrate in its second overtone by a tuning fork of frequency 440 Hz. The speed of sound in air is 330 m/s. End corrections may be neglected. Let  $P_0$  denote the mean pressure at any point in the pipe, and  $\Delta P_0$ the maximum amplitude of pressure variation.

(a) In case of closed organ pipe as fundamental frequency is  $(v/4L)$ and only odd harmonics  are present, second overtone will mean fifth harmonic and so 
                                             $f=\frac{5v}{4L}=440Hz$
      And hence                   $L=\frac{5\times 330}{4\times 440}=\frac{15}{16}m$
         (b) In terms of pressure as at the position of displacement antinode there is pressure node  and vice-versa, the variation of pressure amplitude is standing pressure waves along the length of the column with $x=0$  at this open end will be
                 $p-\Delta p_0\sin kx=\Delta p_0\sin \left(\frac{2\pi }{\lambda }x\right) [Ask=\frac{2\pi }{\lambda }x]$
    Now as for second overtone $L=(5/4)\lambda$, so at the middle.  
                                 $x=\frac{L}{2}=\frac{5}{8}\lambda$
      And hence
                                   $p=\Delta p_0\sin \frac{2\pi }{\lambda }\left(\frac{5}{8}\lambda \right)=\Delta p_0\sin \left(\frac{5}{4}\pi \right)$
      Or                  $\left|p\right|=\Delta p_0\times \frac{1}{\sqrt{2}}=\frac{\Delta p_0}{\sqrt{2}}$
(c) For free end as $x=0,p=0$ , i.e., the amplitude of pressure  wave is zero (as it is a  node), so 
   $p_{\max }=p_{\min }=p_0\pm 0=p_0$
(d) For closed end $x+(5/4)\lambda ,$ so the amplitude of pressure wave 
                    $\left|p\right|=\left|\Delta p_0\sin \frac{2\pi }{\lambda }\left(\frac{5}{4}\lambda \right)\right|=\Delta p_0$
Thus maximum and minimum pressures are given as 
                 $p_{\max }=p_0+\Delta p_{0}and{p}_{\min }=p_0-\Delta p_0$