Identify the type of tube and the wave pattern.

The diagram shows a tube that is closed at one end (bottom, indicated by shading) and open at the other end (top). For a stationary wave in a closed tube, there must be a node (point of zero displacement) at the closed end and an antinode (point of maximum displacement) at the open end.

The diagram shows the following pattern from the closed end to the open end:

• A node at the closed end.
• An antinode.
• Another node.
• An antinode at the open end.

The distance from a node to an adjacent antinode is $\frac{\lambda}{4}$.
The distance from an antinode to an adjacent node is also $\frac{\lambda}{4}$.
Tracing the wave pattern in the diagram:

• From the node at the closed end to the first antinode: $\frac{\lambda}{4}$
• From the first antinode to the next node: $\frac{\lambda}{4}$
• From the second node to the antinode at the open end: $\frac{\lambda}{4}$

Therefore, the total length $L$ of the resonating air column is the sum of these segments:

This corresponds to the first overtone (or third harmonic) for a closed tube.

The wavelength is given as $\lambda = 40 \, cm$.

The calculated length $L = 30 \, cm$ matches option (d).

The final answer is $\text{D. 30cm}$.