A Pipe Closed at One End

solved ClosedPipe

Q. A pipe 20 cm $L = 20$ cm long is closed at one end. The speed of sound in air is 340 m/s $v = 340$ m/s. Which harmonic mode of the pipe is resonantly excited by a 430 Hz $ν = 430$ Hz source?

Inputs

length

L

20 cm

speed

v

340.0 m/s

source

ν

430 Hz

Outputs

harmonic

n

How it's solved

Solved with ClosedPipe

Given

$L = 20$ cm
$v = 340$ m/s
$ν = 430$ Hz

Find $n$ .

Working

$L = 20$ cm $= 0.2$ m
harmonics of a pipe closed at one end(pipe) $ν = \frac{n \cdot v}{4 \cdot L}$
$430 = \frac{n \cdot 340}{4 \cdot 0.2}$
$n = 1.01176471$
$n \approx 1$ to 1 significant figure

Conclusion

A pipe closed at one end sounds only odd multiples of its fundamental, the speed over four times its length; the source matches that fundamental.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :a_pipe_closed_at_one_end do4  kind :numerical5  describes "A pipe %{length} long is closed at one end. The speed of sound in air is %{speed}. %{harmonic} a " \6            "%{source} source?"7  given length: "20 cm"8  given speed: "340 m/s"9  given source: "430 Hz"10  asks harmonic: "Which harmonic mode of the pipe is resonantly excited by"11end

solution.rb

1require_relative "../questions/a_pipe_closed_at_one_end"2require_relative "../science/waves/closed_pipe"34solve :a_pipe_closed_at_one_end, using: ClosedPipe do5  objects :pipe67  take length: pipe.length_of_the_pipe8  take speed: v9  take source: pipe.frequency1011  harmonic = pipe.harmonic12  answer harmonic:, figures: 11314  conclusion "A pipe closed at one end sounds only odd multiples of its fundamental, the speed over four times " \15             "its length; the source matches that fundamental."16end

playground.rb

1require_relative "../solutions/a_pipe_closed_at_one_end"23play_with :a_pipe_closed_at_one_end do4  title "A Pipe Closed at One End"5end

The laws it drew on

quantity	written	units	within	from
speed of sound	$v$	m/s	—	ClosedPipe
length of the pipe	$L$	m	—	ClosedPipe
harmonic	$n$	—	—	ClosedPipe
frequency	$ν$	Hz	0 … ∞	ClosedPipe
frequency of ground	$ν_{g}$	Hz	0 … ∞
harmonic of ground	$n_{g}$	—	—
length of the pipe of ground	$L_{g}$	m	—
frequency of pipe	$ν$	Hz	0 … ∞
harmonic of pipe	$n$	—	—
length of the pipe of pipe	$L$	m	—

ν_{p} = \frac{n_{p} \cdot v}{4 \cdot L_{p}}

harmonics of a pipe closed at one end

ν_{g} = \frac{n_{g} \cdot v}{4 \cdot L_{g}}

harmonics of a pipe closed at one end(ground)

ν = \frac{n \cdot v}{4 \cdot L}

harmonics of a pipe closed at one end(pipe)