Ultrasound and Seismic Waves (Higher)

The same physics of waves that lets you hear a voice also lets a doctor see an unborn baby, a ship measure the depth of the ocean, and a geologist map the inside of the Earth — even though no one can dig more than a few kilometres down. The tools are ultrasound (sound too high for us to hear) and seismic waves (the waves an earthquake sends through the planet). Both work by sending a wave out, watching how it reflects or refracts, and timing how long it takes to return. This Higher-tier lesson, part of Topic P5 (Waves in matter) of OCR Gateway Science A, defines ultrasound and its uses, sets out the distance = speed × time calculation (remembering the wave travels there and back), and explains how P-waves and S-waves reveal the Earth's layered structure and its liquid outer core.

By the end of this lesson you should be able to define ultrasound, describe its uses, carry out echo and depth calculations (correctly halving for the there-and-back path), describe P-waves and S-waves and how they differ, and explain how seismic waves provide evidence for the Earth's layered structure and liquid outer core.

(This entire lesson is Higher-tier material in OCR Gateway Science A.)

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What Is Ultrasound?

Ultrasound is sound with a frequency above $20\,000 \text{ Hz}$20000 Hz ($20 \text{ kHz}$20 kHz) — that is, above the upper limit of human hearing. It is an ordinary longitudinal sound wave in every respect except that its frequency is too high for the human ear to detect. Many animals — bats, dolphins, dogs — can hear and use ultrasound, and we generate it electronically for a range of practical jobs.

The key behaviour that makes ultrasound useful is that, like all waves, it is partly reflected whenever it meets a boundary between two different materials. By sending a pulse of ultrasound into an object and timing how long the reflected pulse (the echo) takes to come back, we can work out how far away the boundary is. This is the principle behind ultrasound scanning, flaw detection and sonar.

Exam Tip: Ultrasound is defined purely by its frequency above $20 \text{ kHz}$20 kHz — the top of the human hearing range. It is still a normal longitudinal sound wave; humans simply cannot hear it.

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Uses of Ultrasound

Medical imaging (foetal scanning). A handheld probe sends pulses of ultrasound into the body. At each boundary between different tissues (and at the surface of a developing baby), some ultrasound is reflected. A computer times the echoes and uses them to build up an image on a screen. Ultrasound is used in preference to X-rays for scanning unborn babies because, unlike X-rays, it is non-ionising and so does not damage living cells — it is considered safe for the foetus.

Industrial flaw detection. Ultrasound is sent into a metal casting, weld or rail. If there is a crack or void inside, the ultrasound reflects off the internal boundary and an extra echo returns earlier than the echo from the far side. This reveals hidden defects without cutting the object open (non-destructive testing).

Sonar and echo sounding. A ship sends a pulse of ultrasound down through the water; it reflects off the sea bed (or off a shoal of fish, or a submarine) and returns. Timing the echo gives the depth of the water. This is echo sounding (sonar).

In every case the method is the same: send a pulse, time the echo, and convert the time into a distance using distance = speed × time — being careful that the pulse travels to the boundary and back.

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The There-and-Back Calculation

This is the calculation OCR tests most often, and the place students most often go wrong. When you send a pulse and time its echo, the wave travels to the boundary and back again — so the distance it covers in that time is twice the distance to the boundary.

The basic relationship is:

$\text{distance} = \text{speed} \times \text{time} \qquad (s = v \, t)$distance=speed×time(s=vt)

But the total distance travelled by the pulse is there and back, which is twice the depth (or distance to the object). So if $d$d is the distance to the boundary:

$2d = v \, t \qquad \Rightarrow \qquad d = \dfrac{v \, t}{2}$2d=vt⇒d=2vt​

In words: multiply the speed by the time to get the total distance, then divide by 2 to get the distance to the boundary. Forgetting to halve gives an answer twice as big as it should be — the classic error.

Worked Example 1 — Depth of the sea (echo sounding)

A ship sends a pulse of ultrasound towards the sea bed. The echo returns $0.4 \text{ s}$0.4 s later. The speed of sound in seawater is $1500 \text{ m/s}$1500 m/s. Calculate the depth of the sea.

Step 1 — find the total distance travelled by the pulse: $\text{distance} = v \times t = 1500 \times 0.4 = 600 \text{ m}$distance=v×t=1500×0.4=600 m.

Step 2 — this is the there-and-back distance, so halve it to find the depth: $d = \dfrac{600}{2} = 300 \text{ m}$d=2600​=300 m.

Answer: the sea is $300 \text{ m}$300 m deep.

Worked Example 2 — A flaw inside metal

Ultrasound is sent into a steel block to look for cracks. An echo returns from a crack after $1.2 \times 10^{-5} \text{ s}$1.2×10−5 s. The speed of ultrasound in steel is $5000 \text{ m/s}$5000 m/s. How far below the surface is the crack?

Step 1 — total distance travelled: $\text{distance} = v \times t = 5000 \times 1.2 \times 10^{-5} = 0.06 \text{ m}$distance=v×t=5000×1.2×10−5=0.06 m.

Step 2 — halve it (there and back): $d = \dfrac{0.06}{2} = 0.03 \text{ m} = 3 \text{ cm}$d=20.06​=0.03 m=3 cm.

Answer: the crack is $0.03 \text{ m}$0.03 m ($3 \text{ cm}$3 cm) below the surface.

Worked Example 3 — Working back to a time

An ultrasound scanner emits a pulse that reflects off a boundary $0.045 \text{ m}$0.045 m deep in the body. Ultrasound travels at $1500 \text{ m/s}$1500 m/s in soft tissue. Calculate the time for the echo to return.

Step 1 — find the total distance (there and back): $2d = 2 \times 0.045 = 0.09 \text{ m}$2d=2×0.045=0.09 m.

Step 2 — rearrange $\text{distance} = v \times t$distance=v×t for time: $t = \dfrac{\text{distance}}{v} = \dfrac{0.09}{1500}$t=vdistance​=15000.09​.

Step 3 — calculate: $t = 6 \times 10^{-5} \text{ s}$t=6×10−5 s.

Answer: the echo returns after $6 \times 10^{-5} \text{ s}$6×10−5 s ($0.00006 \text{ s}$0.00006 s).

Exam Tip: Always ask: did the pulse travel there and back? If you are given the echo time, the wave has covered twice the depth, so either halve the distance at the end, or use $2d$2d from the start. Halving is the single most-tested point in ultrasound calculations — never skip it.

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Seismic Waves: P-waves and S-waves

An earthquake releases energy that travels through the Earth as seismic waves, detected at the surface by instruments called seismometers. There are two main types that travel through the body of the Earth, and they behave very differently:

• P-waves (primary waves) are longitudinal waves (compressions and rarefactions). They are faster and arrive first. Crucially, they can travel through both solids and liquids.
• S-waves (secondary waves) are transverse waves. They are slower and arrive second. Crucially, they can travel through solids only — they cannot pass through liquids.

The reason for the difference lies in the wave type. A longitudinal P-wave squashes and stretches the material along its path, and both solids and liquids resist being compressed, so P-waves pass through both. A transverse S-wave needs the material to be able to "spring back" sideways (resist a shear); solids can do this but liquids cannot, so S-waves are stopped by liquid.