A Steel Wire Under Tension

Q. A steel wire 0.72 m long has a mass of 5.0 × 10⁻³ kg. If the wire is under a tension of 60 N, what is the speed of transverse waves on the wire, in m/s? expects 93 m/s

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

Q. A steel wire 0.72 m $L = 0.72$ m long has a mass of 5.0 × 10⁻³ kg $m = 5.0 \times 10^{−3}$ kg. If the wire is under a tension of 60 N $T = 60$ N, what is the speed of transverse waves on the wire, in m/s?

The playground

Inputs

length

L

0.72 m

mass

m

0.005 kg

tension

T

60 N

Outputs

speed

v

The solution

Solved with WavesOnAString

Given

$L = 0.72$ m
$m = 5.0 \times 10^{−3}$ kg
$T = 60$ N

Find $v$ .

Working

mass per unit length(wire) $μ = \frac{m}{L}$
$μ = \frac{0.005}{0.72}$
$μ = 0.00694444444$ kg/m
speed of a transverse wave(wire) $v = \sqrt{\frac{T}{μ}}$
$v = \sqrt{\frac{60}{0.00694444444}}$
$v = 93.0 m/s$
$v \approx 93$ m/sto 2 significant figures

Conclusion

Tension pulls a disturbance along and the wire's mass per metre holds it back; the speed is the square root of their ratio.

The Ruby

1require_relative "../physics"23question :a_steel_wire_under_tension do4  kind :numerical5  describes "A steel wire %{length} long has a mass of %{mass}. If the wire is under a tension of %{tension}, " \6            "%{speed}, in m/s?"7  given length: "0.72 m"8  given mass: "5.0e-3 kg"9  given tension: "60 N"10  asks speed: "what is the speed of transverse waves on the wire"11end

1require_relative "../questions/a_steel_wire_under_tension"2require_relative "../science/waves/waves_on_a_string"34solve :a_steel_wire_under_tension, using: WavesOnAString do5  objects :wire67  take length: wire.length_of_the_string8  take mass: wire.mass9  take tension: wire.tension1011  speed = wire.wave_speed12  answer speed:, in: "m/s", figures: 21314  conclusion "Tension pulls a disturbance along and the wire's mass per metre holds it back; the speed is " \15             "the square root of their ratio."16end

1require_relative "../solutions/a_steel_wire_under_tension"23play_with :a_steel_wire_under_tension do4  title "A Steel Wire Under Tension"5end

The laws

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)