Derive an expression for the fundamental frequency of one end closed pipe. Discuss the various overtones emitted by the pipe.

Consider, a hollow pipe of length L closed at one end.
A progressive longitudinal wave be sent in the pipe towards left.
The wave gets reflected from closed end.
Both incident and reflected waves superimpose in the pipe and give rise to stationary wave.
Incident and reflected waves are represented by, 
                                ...(1) 
                                 ...(2) 
On superposition, 
                
                  
                
                                                  ...(3) 
which is the resultant stationary wave.
Pipe is open at end B and closed at end A.
Since the air is free to vibrate at open end, therefore open end of the pipe is antinode.
At closed end, the amplitude of vibration is zero and hence closed end is node.
Thus the boundary conditions are: 
At A,  x = 0;       y = 0 
At B,   x = L;       
From equation (3), it follows that 
                      
The solution is trivial. 
 
          
            
                where n is an integer. 
            
                
∴              
where, v is the velocity of wave in the air column.
Fundamental frequency or 1^st mode of vibration i.e. n = 1
 
This mode is the fundamental mode of vibration. 
 
First overtone or 2^nd mode of vibration i.e. n = 2. 
 
The Frequency of second mode of vibration is three times the fundamental frequency.
Hence, it is known as the third harmonic.
3rd mode of vibration i.e. n = 3 
 
Frequency of this mode of vibration is five times the fundamental frequency, hence known as fifth harmonic. 
Similarly, in n^th mode of vibration, the frequency of vibration is,
 
overtone emitted is (n–1)th^.
Since (2n–1) is an odd integer, therefore the overtones emitted by one end closed pipe have only odd harmonics.