An Ultrasonic Scanner

Q. A hospital uses an ultrasonic scanner to locate tumours in a tissue. The operating frequency of the scanner is 4.2 MHz. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km/s, in metres? expects 4.0 × 10⁻⁴ m

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The web of concepts

given read from the words equation asked

The equations it used

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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?

The playground

Inputs

speed

v

1.7 km/s

frequency

ν

4.2 MHz

Outputs

wavelength

λ

The solution

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 Ruby

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

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

1require_relative "../solutions/an_ultrasonic_scanner"23play_with :an_ultrasonic_scanner do4  title "An Ultrasonic Scanner"5end

The laws

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)