Pipe Closed at One End

There is a displacement node at the closed end, and an antinode at the open end. This is just like a string fixed at one end and free at the other. Let's arbitrarily make $x = 0$ the closed end. Then:

$$s(x,t) = s_0 \sin(k_n x )\cos(\omega_n t)$$ (185)

has a node at $x = 0$ for all $k$. To get an antinode at the other end, we require:

$$\sin(k_n L) = \pm 1$$ (186)

or

$$k_n L = \frac{2n-2}{2}\pi$$ (187)

for $n = 1,2,3...$ (odd half-integral multiples of $\pi$. This converts to:

$$\lambda_n = \frac{2L}{2n-1}$$ (188)

and

$$f_n = \frac{v_a}{\lambda_n} = \frac{v_a (2n-1)}{4L}$$ (189)