A Jerk Along a String

Q. A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end, in seconds? expects 0.50 s

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The equations it used

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Check an answer

Q. A string of mass 2.50 kg $m = 2.50$ kg is under a tension of 200 N $T = 200$ N. The length of the stretched string is 20.0 m $L = 20.0$ m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end, in seconds?

The playground

Inputs

mass

m

2.5 kg

tension

T

200 N

length

L

20.00 m

Outputs

time

t

The solution

Solved with WavesOnAString

Given

$m = 2.50$ kg
$T = 200$ N
$L = 20.0$ m

Find $t$ .

Working

mass per unit length(string) $μ = \frac{m}{L}$
$μ = \frac{2.5}{20}$
$μ = 0.125$ kg/m
speed of a transverse wave(string) $v = \sqrt{\frac{T}{μ}}$
$v = \sqrt{\frac{200}{0.125}}$
$v = 40$ m/s
time to run the length(string) $t = \frac{L}{v}$
$t = \frac{20}{40}$
$t = 0.50 s$
$t \approx 0.50$ sto 2 significant figures

Conclusion

The jerk runs along at the speed the tension and the mass per metre set, so it crosses the string in its length over that speed.

The Ruby

1require_relative "../physics"23question :a_jerk_along_a_string do4  kind :numerical5  describes "A string of mass %{mass} is under a tension of %{tension}. The length of the stretched string is " \6            "%{length}. If the transverse jerk is struck at one end of the string, %{time} to reach the other end, " \7            "in seconds?"8  given mass: "2.50 kg"9  given tension: "200 N"10  given length: "20.0 m"11  asks time: "how long does the disturbance take"12end

1require_relative "../questions/a_jerk_along_a_string"2require_relative "../science/waves/waves_on_a_string"34solve :a_jerk_along_a_string, using: WavesOnAString do5  objects :string67  take mass: string.mass8  take tension: string.tension9  take length: string.length_of_the_string1011  time = string.travel_time12  answer time:, in: "s", figures: 21314  conclusion "The jerk runs along at the speed the tension and the mass per metre set, so it crosses the string " \15             "in its length over that speed."16end

1require_relative "../solutions/a_jerk_along_a_string"23play_with :a_jerk_along_a_string do4  title "A Jerk Along a String"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 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)