AS Level Physics 9702: Stationary Waves in a Closed Air Column

A narrow pipe of length $0.75\text{ m}$ is closed at one end and open at the other. A loudspeaker held near the open end emits sound of variable frequency, and the speed of sound in the air inside the pipe is $330\text{ m s}^{-1}$ .

(a) State the condition on the air displacement at the closed end, and at the open end, of the pipe that must be satisfied for a stationary wave to form inside it. [2]

(b) Show that the wavelength of the fundamental (lowest-frequency) stationary wave that can form in this pipe is $3.0\text{ m}$ , and calculate the corresponding fundamental frequency. [3]

(c) Explain why the second harmonic (at twice the fundamental frequency) cannot be produced in this pipe, and state which harmonic is next produced above the fundamental as the frequency is increased. [2]

(d) Calculate the frequency of this next harmonic identified in (c). [2]

Part (a): Boundary conditions

At the closed end, the air molecules are up against a rigid barrier and cannot move, so the closed end must be a displacement node (zero displacement amplitude). At the open end, the air is free to move and vibrates with (approximately) maximum amplitude, so the open end must be a displacement antinode.

Part (b): Fundamental wavelength and frequency

For a pipe closed at one end, the fundamental (lowest) stationary wave has a node at the closed end and an antinode at the open end, with no other nodes or antinodes in between. This is exactly one quarter of a wavelength fitting into the pipe length $L$ : $L = \frac{\lambda_1}{4} \quad\Rightarrow\quad \lambda_1 = 4L$

With $L = 0.75\text{ m}$ : $\lambda_1 = 4 \times 0.75 = 3.0\text{ m}$

as required. Using $v = f\lambda$ , rearranged for frequency: $f_1 = \frac{v}{\lambda_1} = \frac{330}{3.0} = 110\text{ Hz}$

Check (recompute independently): $330 \div 3.0 = 110.0$ exactly, and $110 \times 3.0 = 330$ , consistent with the given speed of sound.

Part (c): Why only odd harmonics occur

Every resonant mode of this pipe must still have a node at the closed end and an antinode at the open end. Fitting an extra half-wavelength each time a new mode is added, the allowed wavelengths are: $L = (2n-1)\frac{\lambda}{4}, \qquad n = 1, 2, 3, \dots$

which correspond to frequencies $f_1, 3f_1, 5f_1, \dots$ , only odd multiples of the fundamental. A “2nd harmonic” pattern ( $2f_1$ ) would require a node at the closed end and a node at the open end (or an antinode at both), which contradicts the fixed boundary condition of node-at-closed-end/antinode-at-open-end. So the second harmonic cannot exist in this pipe, and the next allowed harmonic above the fundamental is the 3rd harmonic.

Part (d): Frequency of the 3rd harmonic

$f_3 = 3 \times f_1 = 3 \times 110 = 330\text{ Hz}$

Check (independent method (via wavelength): the 3rd harmonic has wavelength $\lambda_3 = \dfrac{4L}{3} = \dfrac{3.0}{3} = 1.0\text{ m}$ , so $f_3 = \dfrac{v}{\lambda_3} = \dfrac{330}{1.0} = 330\text{ Hz}$ ) the same result.

Final answers

(a) Closed end: node; open end: antinode
(b) $\lambda_1 = \boxed{3.0\text{ m}}$ ; $f_1 = \boxed{110\text{ Hz}}$
(c) Only odd harmonics occur; the next harmonic is the 3rd harmonic
(d) $f_3 = \boxed{330\text{ Hz}}$