Answer
For a closed pipe which is closed at one end, the only stationary vibrations are those where length of the pipe L is $\frac{2n+1}{4}$ times the wavelength , n=0,1,2,........ since there must be a node at one end and an antinode at the other. The wavelength will be $4L,\frac{4L}{3},\frac{4L}{5},...$
Consequently the harmonics will be 1,3,5 ..... times the fundamental frequency. So, there are no even harmonics.
Work Step by Step
For a closed pipe which is closed at one end, the only stationary vibrations are those where length of the pipe L is $\frac{2n+1}{4}$ times the wavelength , n=0,1,2,........ since there must be a node at one end and an antinode at the other. The wavelength will be $4L,\frac{4L}{3},\frac{4L}{5},...$
Consequently the harmonics will be 1,3,5 ..... times the fundamental frequency. So, there are no even harmonics.
