A Bat's Call into Water

solved TravellingWave

Q. A bat emits ultrasonic sound of frequency 1000 kHz $ν_{c} = 1000$ kHz in air. Speed of sound in water is 1486 m/s $v_{t} = 1486$ m/s. If the sound meets a water surface $ν_{t} = ν_{c}$ , what is the wavelength of the transmitted sound, in metres?

Inputs

frequency

ν_{c}

1000 kHz

speed

v_{t}

1486.0 m/s

Outputs

wavelength

λ_{t}

How it's solved

Solved with TravellingWave

Given

$ν_{c} = 1000$ kHz
$v_{t} = 1486$ m/s
$ν_{t} = ν_{c}$ from “meets a water surface”

Find $λ_{t}$ .

Working

$ν_{c} = 1000$ kHz $= 1 \times 10^{6}$ Hz
meets a water surface $ν_{t} = ν_{c}$
$ν_{t} = 1 \times 10^{6}$
$ν_{t} = 1 \times 10^{6}$ Hz
speed of a wave(transmitted) $v_{t} = ν_{t} \cdot λ_{t}$
$1486 = 1 \times 10^{6} \cdot λ_{t}$
$λ_{t} = 0.001486$ m
$λ_{t} \approx 0.00149$ mto 3 significant figures

Conclusion

Crossing into water the sound keeps the frequency it was made with but runs faster, so each wave stretches longer in the same proportion.

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_bat_s_call_into_water do4  kind :numerical5  describes "A bat emits ultrasonic sound of frequency %{frequency} in air. Speed of sound in water is %{speed}. If " \6            "the sound meets a water surface, %{wavelength}, in metres?"7  given frequency: "1000 kHz"8  given speed: "1486 m/s"9  asks wavelength: "what is the wavelength of the transmitted sound"10end

solution.rb

1require_relative "../questions/a_bat_s_call_into_water"2require_relative "../science/waves/travelling_wave"34solve :a_bat_s_call_into_water, using: TravellingWave do5  objects :call, :transmitted67  take frequency: call.frequency8  take speed: transmitted.wave_speed910  deduce_from "meets a water surface", transmitted.frequency => call.frequency1112  wavelength = transmitted.wavelength13  answer wavelength:, in: "m", figures: 31415  conclusion "Crossing into water the sound keeps the frequency it was made with but runs faster, so each wave " \16             "stretches longer in the same proportion."17end

playground.rb

1require_relative "../solutions/a_bat_s_call_into_water"23play_with :a_bat_s_call_into_water do4  title "A Bat's Call into Water"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 call	$k_{c}$	m⁻¹	—
wavelength of call	$λ_{c}$	m	0 … ∞
wave number of transmitted	$k_{t}$	m⁻¹	—
wavelength of transmitted	$λ_{t}$	m	0 … ∞
angular frequency of ground	$ω_{g}$	rad/s	—
period of ground	$T_{g}$	s	—
angular frequency of call	$ω_{c}$	rad/s	—
period of call	$T_{c}$	s	—
angular frequency of transmitted	$ω_{t}$	rad/s	—
period of transmitted	$T_{t}$	s	—
frequency of ground	$ν_{g}$	Hz	0 … ∞
frequency of call	$ν_{c}$	Hz	0 … ∞
frequency of transmitted	$ν_{t}$	Hz	0 … ∞
wave speed of ground	$v_{g}$	m/s	—
wave speed of call	$v_{c}$	m/s	—
wave speed of transmitted	$v_{t}$	m/s	—
phase difference of ground	$Δ φ_{g}$	—	—
separation of ground	$Δ x_{g}$	m	—
phase difference of call	$Δ φ_{c}$	—	—
separation of call	$Δ x_{c}$	m	—
phase difference of transmitted	$Δ φ_{t}$	—	—
separation of transmitted	$Δ x_{t}$	m	—
amplitude of call	$a_{c}$	m	—
amplitude of transmitted	$a_{t}$	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_{c} = \frac{2 \cdot π}{λ_{c}}

wave number and wavelength(call)

k_{t} = \frac{2 \cdot π}{λ_{t}}

wave number and wavelength(transmitted)

ω_{g} = \frac{2 \cdot π}{T_{g}}

angular frequency and period(ground)

ω_{c} = \frac{2 \cdot π}{T_{c}}

angular frequency and period(call)

ω_{t} = \frac{2 \cdot π}{T_{t}}

angular frequency and period(transmitted)

ν_{g} = \frac{1}{T_{g}}

frequency and period(ground)

ν_{c} = \frac{1}{T_{c}}

frequency and period(call)

ν_{t} = \frac{1}{T_{t}}

frequency and period(transmitted)

v_{g} = ν_{g} \cdot λ_{g}

speed of a wave(ground)

v_{c} = ν_{c} \cdot λ_{c}

speed of a wave(call)

v_{t} = ν_{t} \cdot λ_{t}

speed of a wave(transmitted)

Δ φ_{g} = k_{g} \cdot Δ x_{g}

phase across a separation(ground)

Δ φ_{c} = k_{c} \cdot Δ x_{c}

phase across a separation(call)

Δ φ_{t} = k_{t} \cdot Δ x_{t}

phase across a separation(transmitted)

ν_{t} = ν_{c}

meets a water surface