Between Two Crests

solved TravellingWave

Q. A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x $k = 0.018$ rad/cm + π/4) where x and y are in cm and t in s. What is the least distance between two successive crests in the wave, in metres?

Outputs

distance

λ

How it's solved

Solved with TravellingWave

Given

$k = 0.018$ rad/cm from “0.018 x”

Find $λ$ .

Working

$k = 0.018$ rad/cm $= 1.8$ m⁻¹
wave number and wavelength(wave) $k = \frac{2 \cdot π}{λ}$
$1.8 = \frac{2 \cdot 3.14159265}{λ}$
$λ = 3.49 m$
$λ \approx 3.5$ mto 2 significant figures

Conclusion

Successive crests are a full turn of phase apart along the string, and that distance is the wavelength.

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 :between_two_crests do4  kind :numerical5  describes "A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x + π/4) where x " \6            "and y are in cm and t in s. %{distance}, in metres?"7  asks distance: "What is the least distance between two successive crests in the wave"8end

solution.rb

1require_relative "../questions/between_two_crests"2require_relative "../science/waves/travelling_wave"34solve :between_two_crests, using: TravellingWave do5  objects :wave67  deduce_from "0.018 x", wave.wave_number => "0.018 rad/cm"89  distance = wave.wavelength10  answer distance:, in: "m", figures: 21112  conclusion "Successive crests are a full turn of phase apart along the string, and that distance is the " \13             "wavelength."14end

playground.rb

1require_relative "../solutions/between_two_crests"23play_with :between_two_crests do4  title "Between Two Crests"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	TravellingWave
amplitude	$a$	m	—	TravellingWave
wave number	$k$	m⁻¹	—	TravellingWave
angular frequency	$ω$	rad/s	—	TravellingWave
wavelength	$λ$	m	0 … ∞	TravellingWave
period	$T$	s	—	TravellingWave
frequency	$ν$	Hz	0 … ∞	TravellingWave
wave speed	$v$	m/s	—	TravellingWave
separation	$Δ x$	m	—	TravellingWave
phase difference	$Δ φ$	—	—	TravellingWave
wave number of ground	$k_{g}$	m⁻¹	—
wavelength of ground	$λ_{g}$	m	0 … ∞
wave number of wave	$k$	m⁻¹	—
wavelength of wave	$λ$	m	0 … ∞
angular frequency of ground	$ω_{g}$	rad/s	—
period of ground	$T_{g}$	s	—
angular frequency of wave	$ω$	rad/s	—
period of wave	$T$	s	—
frequency of ground	$ν_{g}$	Hz	0 … ∞
frequency of wave	$ν$	Hz	0 … ∞
wave speed of ground	$v_{g}$	m/s	—
wave speed of wave	$v$	m/s	—
phase difference of ground	$Δ φ_{g}$	—	—
separation of ground	$Δ x_{g}$	m	—
phase difference of wave	$Δ φ$	—	—
separation of wave	$Δ x$	m	—
amplitude of wave	$a$	m	—

k_{w} = \frac{2 \cdot π}{λ_{w}}

wave number and wavelength

ω_{w} = \frac{2 \cdot π}{T_{w}}

angular frequency and period

ν_{w} = \frac{1}{T_{w}}

frequency and period

v_{w} = ν_{w} \cdot λ_{w}

speed of a wave

Δ φ_{w} = k_{w} \cdot Δ x_{w}

phase across a separation

k_{g} = \frac{2 \cdot π}{λ_{g}}

wave number and wavelength(ground)

k = \frac{2 \cdot π}{λ}

wave number and wavelength(wave)

ω_{g} = \frac{2 \cdot π}{T_{g}}

angular frequency and period(ground)

ω = \frac{2 \cdot π}{T}

angular frequency and period(wave)

ν_{g} = \frac{1}{T_{g}}

frequency and period(ground)

ν = \frac{1}{T}

frequency and period(wave)

v_{g} = ν_{g} \cdot λ_{g}

speed of a wave(ground)

v = ν \cdot λ

speed of a wave(wave)

Δ φ_{g} = k_{g} \cdot Δ x_{g}

phase across a separation(ground)

Δ φ = k \cdot Δ x

phase across a separation(wave)