A marine biologist is investigating the ocean floor using a sonar system fitted to a research vessel. The sonar emits ultrasound pulses downward from beneath the ship. Explain how the sonar system is able to determine the depth of the ocean floor, and describe what would happen to the speed and wavelength of the ultrasound pulse if it passed from seawater into a denser layer of sediment on the ocean floor.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
A research vessel uses a sonar system that emits ultrasound pulses downward toward the ocean floor. The speed of ultrasound in seawater is approximately 1500 m/s. The sediment layer on the ocean floor is denser than the seawater above it.
Model answer (5 marks)
The sonar emits a pulse of ultrasound that travels through seawater until it reaches the interface between the water and the ocean floor. At this boundary part of the wave is reflected back towards the vessel. A receiver next to the transmitter detects the reflected pulse and the time interval between emission and detection is measured.
Because the speed of ultrasound in seawater is known (≈1500 m s⁻¹) the distance to the floor can be calculated from the round‑trip time: depth = ½ × speed × time.
If the pulse then enters the denser sediment layer, its speed decreases. The frequency of the wave does not change, so the reduced speed means the wavelength shortens (λ = v / f).
Because the speed of ultrasound in seawater is known (≈1500 m s⁻¹) the distance to the floor can be calculated from the round‑trip time: depth = ½ × speed × time.
If the pulse then enters the denser sediment layer, its speed decreases. The frequency of the wave does not change, so the reduced speed means the wavelength shortens (λ = v / f).
Examiner tips
- Show the sequence: emission → reflection → detection → time measurement → depth calculation.
- Mention the ½ factor for round‑trip distance.
- State that frequency stays constant, so λ changes with v.
- Use the given speed 1500 m s⁻¹ in the calculation.
Mark scheme (5 marks)
- Ultrasound pulses are partially reflected back when they reach the boundary between seawater and the ocean floor (or any boundary between two media).
- A receiver next to the emitter detects the reflected pulses and the time between emission and detection is measured.
- Because the speed of the ultrasound waves is constant (in a given medium), the depth can be calculated from the time taken for the pulse to return.
- When the ultrasound passes into the denser sediment layer, its speed decreases.
- Because frequency remains constant, the decrease in speed causes the wavelength to decrease.
Key terms in this question
Related
- All OCR A-Level Physics A (H556) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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