A pipe open at both ends has a fundamental frequency $f$ in air. The pipe is now dipped vertically in a water drum to half of its length. The fundamental frequency of the air column is now equal to:
A
$\frac{f}{2}$
B
$f$
C
$\frac{3f}{2}$
D
$2f$
The answer is B.
Solution
Initial open pipe has fundamental $f = \frac{v}{2L}$. Half-submerged pipe is closed at one end of length $L/2$, giving fundamental $f' = \frac{v}{4(L/2)} = \frac{v}{2L} = f$.
For an open organ pipe of length $L$, the fundamental frequency is $f = \frac{v}{2L}$. When the pipe is immersed to half of its length in water, the air column length becomes $L' = L/2$, with the water surface acting as a rigid closed end (closed organ pipe). The fundamental frequency of a closed pipe of length $L'$ is $f' = \frac{v}{4 L'} = \frac{v}{4 (L/2)} = \frac{v}{2L} = f$. Therefore, the fundamental frequency remains unchanged and is equal to $f$.
ConceptOpen and Closed Organ Pipes
Waves in past papers
59 questions from this chapter have appeared across 25 exam years.
