Question

A pipe of 1 m length is closed at one end. Taking the speed of sound in air as 320 $\text{ms}^{-1}$, the air column in the pipe cannot resonate for the frequency (in Hz)

• A. 80
• B. 160
• C. 240
• D. 560
• E. 720

Hint:

Pipes closed at one end only support odd harmonics ($1, 3, 5, \dots$), whereas pipes open at both ends support all harmonics ($1, 2, 3, \dots$).

Answer 1

The Correct Option is B

Concept:
For a pipe closed at one end, only odd harmonics are present. The resonance frequencies ($f_n$) are given by: \[ f_n = (2n - 1) \frac{v}{4L} \quad \text{where } n = 1, 2, 3, \ldots \] This means the allowed frequencies are odd multiples of the fundamental frequency ($f_1, 3f_1, 5f_1, \ldots$).

Step 1: Calculate the fundamental frequency ($f_1$).
Given length $L = 1$ m and speed of sound $v = 320$ ms$^{-1}$. \[ f_1 = \frac{v}{4L} = \frac{320}{4 \times 1} = 80 \text{ Hz} \]

Step 2: Identify the allowed harmonics.
The resonance frequencies must be odd multiples of 80 Hz: [itemsep=4pt]
• $f_1 = 1 \times 80 = 80 \text{ Hz}$ (Option A is possible)
• $f_2 = 3 \times 80 = 240 \text{ Hz}$ (Option C is possible)
• $f_3 = 5 \times 80 = 400 \text{ Hz}$
• $f_4 = 7 \times 80 = 560 \text{ Hz}$ (Option D is possible)
• $f_5 = 9 \times 80 = 720 \text{ Hz}$ (Option E is possible) 160 Hz is an even multiple ($2 \times 80$), which is not allowed for a pipe closed at one end.