As the air column of length L in an air pipe decreases, the frequency of the sationary wave emitted

Analyze the question: The question asks how the frequency of a stationary (standing) wave in an air column changes when the length L of the column decreases.

Concept: In a pipe (open or closed), resonance happens when the air column fits a fixed number of wavelengths or fractions of a wavelength.

• Closed pipe, fundamental: L = λ/4, so λ = 4L
• Open pipe, fundamental: L = λ/2, so λ = 2L

The wave speed v is fixed for a given temperature, and v = fλ, so f = v/λ.

Solution:

• Closed pipe: f = v/(4L)
• Open pipe: f = v/(2L)

In both cases f ∝ 1/L, so frequency is inversely proportional to the length of the air column.

When L decreases, 1/L increases, so f increases.

Checking the other options:

• A (decreases): wrong, since f ∝ 1/L means f rises as L falls.
• C (varies as L²): wrong, because the relationship is 1/L, not L².
• D (varies as √L): wrong, for the same reason.

Tip: Shorter air column means higher pitch. This is why a short organ pipe or a short flute produces a shrill note, while a long one gives a deep note. Remember f ∝ 1/L, "shorter pipe, higher frequency."