A Clamped String's Tension

solved WavesOnAString

Q. The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin (2π/3 x) $k = 120$ deg/m cos (120 πt) $ω = 60$ rev/s where x and y are in m and t in s. The length of the string is 1.5 m $L = 1.5$ m and its mass is 3.0 × 10⁻² kg $m = 3.0 \times 10^{−2}$ kg. Determine the tension in the string, in newtons?

Inputs

length

L

1.50 m

mass

m

0.03 kg

Outputs

tension

T

How it's solved

Solved with WavesOnAString

Given

$L = 1.5$ m
$m = 3.0 \times 10^{−2}$ kg
$k = 120$ deg/m from “sin (2π/3 x)”
$ω = 60$ rev/s from “cos (120 πt)”

Find $T$ .

Working

$k = 120$ deg/m $= 2.0943951$ m⁻¹
$ω = 60$ rev/s $= 376.991118$ rad/s
speed from the wave s numbers(string) $v = \frac{ω}{k}$
$v = \frac{376.991118}{2.0943951}$
$v = 180$ m/s
mass per unit length(string) $μ = \frac{m}{L}$
$μ = \frac{0.03}{1.5}$
$μ = 0.02$ kg/m
speed of a transverse wave(string) $v = \sqrt{\frac{T}{μ}}$
$180 = \sqrt{\frac{T}{0.02}}$
$T = 648$ N
$T \approx 648$ Nto 3 significant figures

Conclusion

The two waves that make up the standing wave run at the speed the tension sets against the mass per metre, so that speed squared times the mass per metre is the tension.

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_clamped_string_s_tension do4  kind :numerical5  describes "The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin (2π/3 x) " \6            "cos (120 πt) where x and y are in m and t in s. The length of the string is %{length} and its mass is " \7            "%{mass}. %{tension}, in newtons?"8  given length: "1.5 m"9  given mass: "3.0e-2 kg"10  asks tension: "Determine the tension in the string"11end

solution.rb

1require_relative "../questions/a_clamped_string_s_tension"2require_relative "../science/waves/waves_on_a_string"34solve :a_clamped_string_s_tension, using: WavesOnAString do5  objects :string67  take length: string.length_of_the_string8  take mass: string.mass910  deduce_from "sin (2π/3 x)", string.wave_number => "120 deg/m"11  deduce_from "cos (120 πt)", string.angular_frequency => "60 rev/s"1213  tension = string.tension14  answer tension:, in: "N", figures: 31516  conclusion "The two waves that make up the standing wave run at the speed the tension sets against the mass per " \17             "metre, so that speed squared times the mass per metre is the tension."18end

playground.rb

1require_relative "../solutions/a_clamped_string_s_tension"23play_with :a_clamped_string_s_tension do4  title "A Clamped String's Tension"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 string	$μ$	kg/m	—
mass of string	$m$	kg	0 … ∞
length of the string of string	$L$	m	—
wave speed of ground	$v_{g}$	m/s	—
tension of ground	$T_{g}$	N	—
wave speed of string	$v$	m/s	—
tension of string	$T$	N	—
angular frequency of ground	$ω_{g}$	rad/s	—
wave number of ground	$k_{g}$	m⁻¹	—
angular frequency of string	$ω$	rad/s	—
wave number of string	$k$	m⁻¹	—
travel time of ground	$t_{g}$	s	—
travel time of string	$t$	s	—
frequency of ground	$ν_{g}$	Hz	0 … ∞
harmonic of ground	$n_{g}$	—	—
frequency of string	$ν$	Hz	0 … ∞
harmonic of string	$n$	—	—

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(string)

v_{g} = \sqrt{\frac{T_{g}}{μ_{g}}}

speed of a transverse wave(ground)

v = \sqrt{\frac{T}{μ}}

speed of a transverse wave(string)

v_{g} = \frac{ω_{g}}{k_{g}}

speed from the wave s numbers(ground)

v = \frac{ω}{k}

speed from the wave s numbers(string)

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

time to run the length(ground)

t = \frac{L}{v}

time to run the length(string)

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(string)