A Steel Wire as Fast as Sound

solved WavesOnAString

Q. A steel wire has a length of 12.0 m $L = 12.0$ m and a mass of 2.10 kg $m = 2.10$ kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m/s $v = 343$ m/s, in newtons?

Inputs

length

L

12.00 m

mass

m

2.1 kg

speed

v

343.0 m/s

Outputs

tension

T

How it's solved

Solved with WavesOnAString

Given

$L = 12.0$ m
$m = 2.10$ kg
$v = 343$ m/s

Find $T$ .

Working

mass per unit length(wire) $μ = \frac{m}{L}$
$μ = \frac{2.1}{12}$
$μ = 0.175$ kg/m
speed of a transverse wave(wire) $v = \sqrt{\frac{T}{μ}}$
$343 = \sqrt{\frac{T}{0.175}}$
$T = 20590$ N
$T \approx 2.06 \times 10^{4}$ Nto 3 significant figures

Conclusion

The speed goes as the square root of the tension over the mass per metre, so matching the speed of sound takes the mass per metre times that speed squared.

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_steel_wire_as_fast_as_sound do4  kind :numerical5  describes "A steel wire has a length of %{length} and a mass of %{mass}. %{tension} so that speed of a transverse " \6            "wave on the wire equals the speed of sound in dry air at 20 °C = %{speed}, in newtons?"7  given length: "12.0 m"8  given mass: "2.10 kg"9  given speed: "343 m/s"10  asks tension: "What should be the tension in the wire"11end

solution.rb

1require_relative "../questions/a_steel_wire_as_fast_as_sound"2require_relative "../science/waves/waves_on_a_string"34solve :a_steel_wire_as_fast_as_sound, using: WavesOnAString do5  objects :wire67  take length: wire.length_of_the_string8  take mass: wire.mass9  take speed: wire.wave_speed1011  tension = wire.tension12  answer tension:, in: "N", figures: 31314  conclusion "The speed goes as the square root of the tension over the mass per metre, so matching the speed " \15             "of sound takes the mass per metre times that speed squared."16end

playground.rb

1require_relative "../solutions/a_steel_wire_as_fast_as_sound"23play_with :a_steel_wire_as_fast_as_sound do4  title "A Steel Wire as Fast as Sound"5end

The laws it drew on

quantity	written	units	within	from
mass	$m$	kg	0 … ∞	WavesOnAString
length of the string	$L$	m	—	WavesOnAString
linear mass density	$μ$	kg/m	—	WavesOnAString
tension	$T$	N	—	WavesOnAString
wave speed	$v$	m/s	—	WavesOnAString
travel time	$t$	s	—	WavesOnAString
harmonic	$n$	—	—	WavesOnAString
frequency	$ν$	Hz	0 … ∞	WavesOnAString
wave number	$k$	m⁻¹	—	WavesOnAString
angular frequency	$ω$	rad/s	—	WavesOnAString
linear mass density of ground	$μ_{g}$	kg/m	—
mass of ground	$m_{g}$	kg	0 … ∞
length of the string of ground	$L_{g}$	m	—
linear mass density of wire	$μ$	kg/m	—
mass of wire	$m$	kg	0 … ∞
length of the string of wire	$L$	m	—
wave speed of ground	$v_{g}$	m/s	—
tension of ground	$T_{g}$	N	—
wave speed of wire	$v$	m/s	—
tension of wire	$T$	N	—
angular frequency of ground	$ω_{g}$	rad/s	—
wave number of ground	$k_{g}$	m⁻¹	—
angular frequency of wire	$ω$	rad/s	—
wave number of wire	$k$	m⁻¹	—
travel time of ground	$t_{g}$	s	—
travel time of wire	$t$	s	—
frequency of ground	$ν_{g}$	Hz	0 … ∞
harmonic of ground	$n_{g}$	—	—
frequency of wire	$ν$	Hz	0 … ∞
harmonic of wire	$n$	—	—

μ_{s} = \frac{m_{s}}{L_{s}}

mass per unit length

v_{s} = \sqrt{\frac{T_{s}}{μ_{s}}}

speed of a transverse wave

v_{s} = \frac{ω_{s}}{k_{s}}

speed from the wave s numbers

t_{s} = \frac{L_{s}}{v_{s}}

time to run the length

ν_{s} = \frac{n_{s} \cdot v_{s}}{2 \cdot L_{s}}

harmonics of a string fixed at both ends

μ_{g} = \frac{m_{g}}{L_{g}}

mass per unit length(ground)

μ = \frac{m}{L}

mass per unit length(wire)

v_{g} = \sqrt{\frac{T_{g}}{μ_{g}}}

speed of a transverse wave(ground)

v = \sqrt{\frac{T}{μ}}

speed of a transverse wave(wire)

v_{g} = \frac{ω_{g}}{k_{g}}

speed from the wave s numbers(ground)

v = \frac{ω}{k}

speed from the wave s numbers(wire)

t_{g} = \frac{L_{g}}{v_{g}}

time to run the length(ground)

t = \frac{L}{v}

time to run the length(wire)

ν_{g} = \frac{n_{g} \cdot v_{g}}{2 \cdot L_{g}}

harmonics of a string fixed at both ends(ground)

ν = \frac{n \cdot v}{2 \cdot L}

harmonics of a string fixed at both ends(wire)