Organ pipe harmonics open at both ends produce all integer multiples of the fundamental frequency, like 50Hz, 100Hz, 150Hz, and 200Hz. But closing one end transforms the standing wave pattern, retaining only odd harmonics of a new lower fundamental. This CSIR NET-level question tests understanding of boundary conditions in sound waves.

Open Pipe Harmonics

An organ pipe open at both ends supports standing waves with antinodes at both ends, producing all harmonics where frequencies follow \( f_n = n \cdot f_1 \), and \( f_1 = \frac{v}{2L} \). Here, the first four harmonics are 50 Hz (n=1), 100 Hz (n=2), 150 Hz (n=3), and 200 Hz (n=4), confirming \( f_1 = 50 \) Hz and speed-to-length ratio \( \frac{v}{L} = 100 \) Hz.

Closed Pipe Harmonics

Closing one end creates a node at the closed end and an antinode at the open end, allowing only odd harmonics: \( f_m = (2m-1) \cdot f_1′ \), where \( f_1′ = \frac{v}{4L} = 25 \) Hz (half the open fundamental due to quarter-wavelength). Possible frequencies are 25 Hz (m=1), 75 Hz (m=2), 125 Hz (m=3), 175 Hz (m=4), 225 Hz (m=5), and so on.

Option Analysis

• 50 Hz and 150 Hz only: Incorrect; 50 Hz (even multiple of 25 Hz) and 150 Hz (6×25 Hz, even) do not match odd multiples (2m-1).
• 100 Hz and 200 Hz only: Incorrect; both even harmonics from open pipe, absent in closed pipe.
• 150 Hz and 200 Hz only: Incorrect; neither fits closed pipe’s odd series.
• None: Correct; no original frequencies (50, 100, 150, 200 Hz) appear in the closed pipe’s 25, 75, 125 Hz sequence.

Fundamental Principles of Organ Pipe Harmonics

In pipes open at both ends, antinodes form at both openings, fitting \( L = \frac{n\lambda}{2} \), so \( f_n = n \frac{v}{2L} \) for n=1,2,3,… The given 50Hz (1st), 100Hz (2nd), 150Hz (3rd), 200Hz (4th) confirm this arithmetic progression.

Closing one end introduces a node, changing to \( L = \frac{(2m-1)\lambda}{4} \), yielding \( f_m = (2m-1) \frac{v}{4L} \). With same L and v, the new fundamental halves to 25Hz, followed by 75Hz, 125Hz, etc.—only odd multiples.

Why No Harmonics Remain: Step-by-Step

1. Open pipe: \( f_1 = 50 \) Hz, so \( \frac{v}{2L} = 50 \), \( \frac{v}{L} = 100 \).
2. Closed pipe: \( f_1′ = \frac{v}{4L} = 25 \) Hz.
3. Check originals: 50=2×25 (even), 100=4×25 (even), 150=6×25 (even), 200=8×25 (even)—none odd.

CSIR NET Exam Insights

Such questions appear in waves and oscillations sections, emphasizing boundary effects. Even harmonics vanish; new ones like 75Hz emerge but aren’t options. Practice identifies pipe type from frequency ratios (1:2:3… open; 1:3:5… closed).