An Ultrasonic Scanner

solved TravellingWave

Q. A hospital uses an ultrasonic scanner to locate tumours in a tissue. The operating frequency of the scanner is 4.2 MHz $ν = 4.2$ MHz. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km/s $v = 1.7$ km/s, in metres?

Inputs

speed

v

1.7 km/s

frequency

ν

4.2 MHz

Outputs

wavelength

λ

How it's solved

Solved with TravellingWave

Given

$v = 1.7$ km/s
$ν = 4.2$ MHz

Find $λ$ .

Working

$v = 1.7$ km/s $= 1700$ m/s
$ν = 4.2$ MHz $= 4.2 \times 10^{6}$ Hz
speed of a wave(sound) $v = ν \cdot λ$
$1700 = 4.2 \times 10^{6} \cdot λ$
$λ = 4.048 \times 10^{−4}$ m
$λ \approx 4.0 \times 10^{−4}$ mto 2 significant figures

Conclusion

Each wave travels one wavelength in one period, so the wavelength is the speed in the tissue over the frequency of the scanner.

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 :an_ultrasonic_scanner do4  kind :numerical5  describes "A hospital uses an ultrasonic scanner to locate tumours in a tissue. The operating frequency of the " \6            "scanner is %{frequency}. %{wavelength} in which the speed of sound is %{speed}, in metres?"7  given speed: "1.7 km/s"8  given frequency: "4.2 MHz"9  asks wavelength: "What is the wavelength of sound in the tissue"10end

solution.rb

1require_relative "../questions/an_ultrasonic_scanner"2require_relative "../science/waves/travelling_wave"34solve :an_ultrasonic_scanner, using: TravellingWave do5  objects :sound67  take speed: sound.wave_speed8  take frequency: sound.frequency910  wavelength = sound.wavelength11  answer wavelength:, in: "m", figures: 21213  conclusion "Each wave travels one wavelength in one period, so the wavelength is the speed in the tissue over " \14             "the frequency of the scanner."15end

playground.rb

1require_relative "../solutions/an_ultrasonic_scanner"23play_with :an_ultrasonic_scanner do4  title "An Ultrasonic Scanner"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 sound	$k$	m⁻¹	—
wavelength of sound	$λ$	m	0 … ∞
angular frequency of ground	$ω_{g}$	rad/s	—
period of ground	$T_{g}$	s	—
angular frequency of sound	$ω$	rad/s	—
period of sound	$T$	s	—
frequency of ground	$ν_{g}$	Hz	0 … ∞
frequency of sound	$ν$	Hz	0 … ∞
wave speed of ground	$v_{g}$	m/s	—
wave speed of sound	$v$	m/s	—
phase difference of ground	$Δ φ_{g}$	—	—
separation of ground	$Δ x_{g}$	m	—
phase difference of sound	$Δ φ$	—	—
separation of sound	$Δ x$	m	—
amplitude of sound	$a$	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(sound)

ω_{g} = \frac{2 \cdot π}{T_{g}}

angular frequency and period(ground)

ω = \frac{2 \cdot π}{T}

angular frequency and period(sound)

ν_{g} = \frac{1}{T_{g}}

frequency and period(ground)

ν = \frac{1}{T}

frequency and period(sound)

v_{g} = ν_{g} \cdot λ_{g}

speed of a wave(ground)

v = ν \cdot λ

speed of a wave(sound)

Δ φ_{g} = k_{g} \cdot Δ x_{g}

phase across a separation(ground)

Δ φ = k \cdot Δ x

phase across a separation(sound)