Two Points Half a Metre Apart

solved TravellingWave

Q. For the travelling harmonic wave y(x, t) = 2.0 cos 2π (10t – 0.0080 x $k = 0.0080$ rev/cm + 0.35) where x and y are in cm and t in s, what is the phase difference between oscillatory motion of two points separated by a distance of 0.5 m $Δ x = 0.5$ m, in radians?

Inputs

separation

Δ x

0.50 m

Outputs

phase

Δ φ

How it's solved

Solved with TravellingWave

Given

$Δ x = 0.5$ m
$k = 0.0080$ rev/cm from “2π (10t – 0.0080 x”

Find $Δ φ$ .

Working

$k = 0.0080$ rev/cm $= 5.02654825$ m⁻¹
phase across a separation(wave) $Δ φ = k \cdot Δ x$
$Δ φ = 5.02654825 \cdot 0.5$
$Δ φ = 2.513$ rad
$Δ φ \approx 2.5$ radto 2 significant figures

Conclusion

Along the wave the phase falls behind by the wave number for every unit of distance, so two points differ in phase by the wave number times their separation.

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 :two_points_half_a_metre_apart do4  kind :numerical5  describes "For the travelling harmonic wave y(x, t) = 2.0 cos 2π (10t – 0.0080 x + 0.35) where x and y are in cm " \6            "and t in s, %{phase} separated by a distance of %{separation}, in radians?"7  given separation: "0.5 m"8  asks phase: "what is the phase difference between oscillatory motion of two points"9end

solution.rb

1require_relative "../questions/two_points_half_a_metre_apart"2require_relative "../science/waves/travelling_wave"34solve :two_points_half_a_metre_apart, using: TravellingWave do5  objects :wave67  take separation: wave.separation89  deduce_from "2π (10t – 0.0080 x", wave.wave_number => "0.0080 rev/cm"1011  phase = wave.phase_difference12  answer phase:, in: "rad", figures: 21314  conclusion "Along the wave the phase falls behind by the wave number for every unit of distance, so two points " \15             "differ in phase by the wave number times their separation."16end

playground.rb

1require_relative "../solutions/two_points_half_a_metre_apart"23play_with :two_points_half_a_metre_apart do4  title "Two Points Half a Metre Apart"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)