Statement-1: In an open organ pipe, all harmonics (odd and even multiples of the fundamental frequency) are present.
Statement-2: The fundamental frequency of a closed organ pipe is half the fundamental frequency of an open organ pipe of the same length.

Question

Statement-1: In an open organ pipe, all harmonics (odd and even multiples of the fundamental frequency) are present.

Statement-2: The fundamental frequency of a closed organ pipe is half the fundamental frequency of an open organ pipe of the same length.

TestHub · Accepted Answer

Statement-1 is true. An open organ pipe has antinodes at both ends. The allowed wavelengths are $2 L / n$ , where $n = 1, 2, 3, \dots$ . The frequencies are $f_{n} = n (v /2 L)$ , so all harmonics (fundamental, 2nd, 3rd, etc.) are present.

Statement-2 is false. For an open organ pipe of length $L$ , the fundamental frequency is $f_{o p e n} = v / (2 L)$ . For a closed organ pipe of length $L$ , the fundamental frequency is $f_{c l ose d} = v / (4 L)$ .

Therefore, $f_{c l ose d} = (1/2) f_{o p e n}$ is incorrect. It should be $f_{c l ose d} = (1/2) f_{o p e n}$ if the statement meant $f_{c l ose d} = (1/2) f_{o p e n}$ which is $v / (4 L) = (1/2) v / (2 L)$ which is true. Let me re-read the statement carefully. "The fundamental frequency of a closed organ pipe is half the fundamental frequency of an open organ pipe of the same length." This is $v / (4 L) = (1/2) \times v / (2 L)$ , which simplifies to $v / (4 L) = v / (4 L)$ . So, Statement-2 is actually true. My initial check was wrong. Let me re-evaluate the options. Re-evaluation: $f_{o p e n} = v / (2 L)$ and $f_{c l ose d} = v / (4 L)$ . So, $f_{c l ose d} = (1/2) f_{o p e n}$ . Statement-2 is true. Both statements are true. Statement-2 is not an explanation for Statement-1. So the answer should be B. Final check: Statement-1 is true. Statement-2 is true. They are independent facts. So B is correct.

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