A Wire in Its Fundamental Mode

solved WavesOnAString

Q. A wire stretched between two rigid supports vibrates in its fundamental mode $n = 1$ with a frequency of 45 Hz $ν = 45$ Hz. The mass of the wire is 3.5 × 10⁻² kg $m = 3.5 \times 10^{−2}$ kg and its linear mass density is 4.0 × 10⁻² kg/m $μ = 4.0 \times 10^{−2}$ kg/m. What is the speed of a transverse wave on the string, in m/s?

Inputs

frequency

ν

45 Hz

mass

m

0.035 kg

density

μ

0.04 kg/m

Outputs

speed

v

How it's solved

Solved with WavesOnAString

Given

$ν = 45$ Hz
$m = 3.5 \times 10^{−2}$ kg
$μ = 4.0 \times 10^{−2}$ kg/m
$n = 1$ from “fundamental mode”

Find $v$ .

Working

mass per unit length(wire) $μ = \frac{m}{L}$
$0.04 = \frac{0.035}{L}$
$L = 0.875$ m
harmonics of a string fixed at both ends(wire) $ν = \frac{n \cdot v}{2 \cdot L}$
$45 = \frac{1 \cdot v}{2 \cdot 0.875}$
$v = 78.8 m/s$
$v \approx 79$ m/sto 2 significant figures

Conclusion

The mass and the mass per metre give the wire's length; in the fundamental mode that length is half a wavelength, and the wave runs one wavelength each period.

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_wire_in_its_fundamental_mode do4  kind :numerical5  describes "A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of " \6            "%{frequency}. The mass of the wire is %{mass} and its linear mass density is %{density}. %{speed}, " \7            "in m/s?"8  given frequency: "45 Hz"9  given mass: "3.5e-2 kg"10  given density: "4.0e-2 kg/m"11  asks speed: "What is the speed of a transverse wave on the string"12end

solution.rb

1require_relative "../questions/a_wire_in_its_fundamental_mode"2require_relative "../science/waves/waves_on_a_string"34solve :a_wire_in_its_fundamental_mode, using: WavesOnAString do5  objects :wire67  take frequency: wire.frequency8  take mass: wire.mass9  take density: wire.linear_mass_density1011  deduce_from "fundamental mode", wire.harmonic => 11213  speed = wire.wave_speed14  answer speed:, in: "m/s", figures: 21516  conclusion "The mass and the mass per metre give the wire's length; in the fundamental mode that length is " \17             "half a wavelength, and the wave runs one wavelength each period."18end

playground.rb

1require_relative "../solutions/a_wire_in_its_fundamental_mode"23play_with :a_wire_in_its_fundamental_mode do4  title "A Wire in Its Fundamental Mode"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)