Explanation

(a) Wavelength of closed organ pipe is
$\lambda = \frac{4L}{(2n - 1)}$
Putting n=1,2,3,….. we find that
$\lambda_{1} = 4L,\frac{4L}{3},\frac{4L}{5},\ldots.$
So frequency of vibration corresponding to modes
n=1,2,3…is
$v_{1} = \frac{v}{\lambda_{1}} = \frac{v}{4L} = v_{1}\ $
$v_{2} = \frac{v}{\lambda_{2}} = \frac{v}{4L/3} = \frac{3v}{4L} = 3v_{1}$
$v_{2} = \frac{v}{\lambda_{3}} = \frac{v}{4L/5} = \frac{5v}{4L} = 5v_{1}$
$\therefore\ \ v_{1}Lv_{2}:v_{3}\ldots. = 1:3:5:\ldots.$
So, only odd harmonics are present.