AS Level Physics 9702: Stationary Waves in a Pipe Open at Both Ends

A pipe of length $0.85\text{ m}$ is open at both ends. A loudspeaker held near one end emits sound of variable frequency, and the speed of sound in the air inside the pipe is $340\text{ m s}^{-1}$ .

(a) State the condition on the air displacement that must be satisfied at each open end of the pipe for a stationary wave to form inside it. [1]

(b) Show that the wavelength of the fundamental (lowest-frequency) stationary wave that can form in this pipe is $1.70\text{ m}$ , and calculate the corresponding fundamental frequency. [3]

(d) State the number of displacement nodes present along the pipe (not counting the ends) when the pipe is vibrating at the third harmonic. [1]

Part (a): Boundary condition

At each open end of the pipe, the air is free to vibrate and does so with (approximately) maximum amplitude, so each open end must be a displacement antinode.

Part (b): Fundamental wavelength and frequency

For a pipe open at both ends, the fundamental (lowest) stationary wave has a displacement antinode at each end and a single node in the middle, with no other nodes or antinodes in between. This is exactly half a wavelength fitting into the pipe length $L$ : $L = \frac{\lambda_1}{2} \quad\Rightarrow\quad \lambda_1 = 2L$

With $L = 0.85\text{ m}$ : $\lambda_1 = 2 \times 0.85 = 1.70\text{ m}$

as required. Using $v = f\lambda$ , rearranged for frequency: $f_1 = \frac{v}{\lambda_1} = \frac{340}{1.70} = 200\text{ Hz}$

Check (recompute independently): $200 \times 1.70 = 340$ , matches the given speed of sound exactly.

Part (c): Frequency of the third harmonic

Unlike a pipe closed at one end, a pipe open at both ends supports all integer harmonics (1st, 2nd, 3rd,…), because each successive mode simply adds one more half-wavelength between two antinode ends. So the $n$ th harmonic has frequency $f_n = n f_1$ , and: $f_3 = 3 \times f_1 = 3 \times 200 = 600\text{ Hz}$

Check (independent method (via wavelength): the third harmonic has wavelength $\lambda_3 = \dfrac{2L}{3} = \dfrac{1.70}{3} = 0.5\overline{6}\text{ m}$ , so $f_3 = \dfrac{v}{\lambda_3} = \dfrac{340}{0.5\overline{6}} = 600\text{ Hz}$ ) the same result.

Part (d): Number of nodes in the third harmonic

For the $n$ th harmonic of an open-open pipe, the pipe length fits $n$ half-wavelength “loops” in a row: $L = n\dfrac{\lambda_n}{2}$ . Each loop begins and ends with a displacement antinode, and adjacent loops share their boundary, so there are $n+1$ antinodes in total (the two open ends plus $n-1$ internal ones) and $n$ nodes (one at the centre of each loop).

For the third harmonic ( $n=3$ ), the pattern along the pipe is: antinode (open end) – node – antinode – node – antinode – node – antinode (open end). This has $4$ antinodes and $3$ nodes.

Check (independent method): the node-to-node (and antinode-to-antinode) spacing is always $\lambda_3/2 = 0.5\overline{6}/2 = 0.28\overline{3}\text{ m}$ . Three loops of this length span $3\times0.28\overline{3} = 0.85\text{ m} = L$ , confirming three loops (and hence three internal nodes) fit exactly along the pipe.

So the number of nodes is $\boxed{3}$ .

Final answers

(a) Each open end must be a displacement antinode
(b) $\lambda_1 = \boxed{1.70\text{ m}}$ ; $f_1 = \boxed{200\text{ Hz}}$
(c) $f_3 = \boxed{600\text{ Hz}}$
(d) Number of nodes $= \boxed{3}$