Q139 / 150 · MHT CET 2025 (22 Apr Shift 1) · Physics
An organ pipe has fundamental frequency 80 Hz . If its one end is closed, the frequencies produced will be (in Hz ) (Neglect end correction)
- A$40,80,120,160, \ldots \ldots . . . .$.
- B$40,80,160,240$,
- Correct answerC$40,120,200,280, \ldots \ldots \ldots . .$.
- D$80,160,240,320, \ldots \ldots . . . .$.
Solution
An organ pipe of length $L$ with fundamental frequency $80\text{ Hz}$ is initially open, satisfying $\frac{v}{2L} = 80$, where $v$ is the speed of sound.
When one end is closed, it becomes a closed pipe with fundamental frequency $f_{closed} = \frac{v}{4L} = \frac{1}{2} \cdot 80 = 40\text{ Hz}$.
Closed pipes produce only odd harmonics: $f_{closed}, 3f_{closed}, 5f_{closed}, 7f_{closed}, \ldots$
Substituting $40\text{ Hz}$ yields the sequence $40, 120, 200, 280, \ldots\text{ Hz}$, which corresponds to option C.
Hence, the correct answer is option: 'C'.