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Physics: Waves, Sound, and Light - Ultrasound, Doppler Effect, and Medical Imaging

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Ultrasound is defined as sound waves with frequencies higher than the upper audible limit of human hearing, which is typically 20,000 Hz20,000 \text{ Hz}.

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In medical imaging (sonography), ultrasound pulses are sent into the body and reflect off boundaries between different tissues. The time delay of the echo is used to calculate the depth of the organ using the formula d=v×t2d = \frac{v \times t}{2}.

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Ultrasound is preferred over X-rays for prenatal scanning because it is non-ionizing, meaning it does not damage living cells or DNA.

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The Doppler Effect is the change in the observed frequency (or pitch) of a wave when there is relative motion between the source and the observer.

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If a sound source moves toward an observer, the wavefronts are compressed, leading to a higher frequency (fobs>fsourcef_{obs} > f_{source}). If it moves away, the wavefronts are stretched, leading to a lower frequency.

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In medicine, the Doppler Effect is used in 'Doppler Ultrasound' to measure the speed and direction of blood flow through arteries and veins.

📐Formulae

v=fλv = f \lambda

d=v×t2d = \frac{v \times t}{2}

f=1Tf = \frac{1}{T}

fobs=fs(vv±vs)f_{obs} = f_s \left( \frac{v}{v \pm v_s} \right)

💡Examples

Problem 1:

A sonar system on a ship sends an ultrasound pulse to the bottom of the ocean. The speed of sound in seawater is 1500 m/s1500 \text{ m/s}. If the echo is detected after 3.2 s3.2 \text{ s}, calculate the depth of the ocean floor.

Solution:

d=1500×3.22d = \frac{1500 \times 3.2}{2} d=48002d = \frac{4800}{2} d=2400 md = 2400 \text{ m}

Explanation:

Because the sound wave travels to the bottom and back, the total distance covered is 2d2d. To find the depth (dd), we multiply the speed by the total time and divide by 22.

Problem 2:

An ambulance siren emits a sound at a frequency of 500 Hz500 \text{ Hz}. If the ambulance is moving rapidly towards a stationary observer, will the observer hear a frequency higher than, lower than, or equal to 500 Hz500 \text{ Hz}? Explain using the Doppler Effect.

Solution:

The observer will hear a frequency higher than 500 Hz500 \text{ Hz}.

Explanation:

As the source moves toward the observer, the sound waves in front of the ambulance are 'bunched up' or compressed. This decreases the wavelength (λ\lambda), and since the speed of sound remains constant, the frequency (ff) must increase (v=fλv = f \lambda).

Problem 3:

Calculate the total distance traveled by an ultrasound pulse in 0.0004 s0.0004 \text{ s} if the speed of sound in human tissue is 1540 m/s1540 \text{ m/s}.

Solution:

1540×0.00040.616\begin{array}{r} 1540 \\ \times 0.0004 \\ \hline 0.616 \end{array} Total distance = 0.616 m0.616 \text{ m} or 61.6 cm61.6 \text{ cm}.

Explanation:

To find the total distance (there and back), we use Distance=Speed×TimeDistance = Speed \times Time. Unlike the depth calculation, we do not divide by 22 here because the question asks for the 'total distance traveled'.