Calculate sonar depth and distinguish total path from distance
Use underwater ultrasound echoes and account for round-trip time, supplied water speed and known delays.
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028. Follow the selected spectrum or sound section. Diagrams illustrate stated ideal models and supplied data. Written practice is self-checked. Human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Describe an echo and convert time units.
- Use water sound speed supplied by a question.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What physical event produces the received sonar return?
What you will learn
- Describe sonar using ultrasound pulses and echoes.
- Calculate depth or distance from time and wave speed in a stated geometry.
Trace an underwater reflected route
Sonar can transmit ultrasound into water and detect echoes from the seabed or an object. For a vertical out-and-back pulse with transmitter and receiver effectively at the same point, the travelled path is twice the depth. A supplied 1500 m/s water speed and 0.12 s echo delay give depth 90 m.
Use the speed for the water conditions, not the air value. A slanted path gives a range along that path rather than vertical depth unless geometry is supplied. A moving target or vessel can require additional information beyond the simple stationary model.
Correct the time and reference
If a recorded 0.14 s interval includes a known 0.02 s instrument delay, the sound round-trip time is 0.12 s. Multiply this corrected interval by the stated speed, then halve for the direct return geometry. Known delays are not part of the wave's water travel.
The result measures distance from the transducer position to the model reflector. If the transducer sits below the water surface, a question must specify whether it asks for distance below the transducer or total depth below the surface. State the reference rather than silently adding an unstated offset.
Pause and explain
At 1500 m/s, a 0.12 s direct round-trip echo gives what distance below the transducer?
Worked example
For a vertical direct return, an echo takes 0.12 s in water at 1500 m/s. Find depth below the transducer.
Show the worked solution
- Total sound path = vt = 1500 × 0.12 = 180 m.
- The path includes downward and upward travel.
- Depth = 180/2 = 90 m.
Answer 90 m below the transducer.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find depth below the transducer for a 0.20 s direct echo at 1500 m/s.
Give me a hint
Multiply by speed, then halve the total path.
Compare my reasoning
- Total path = 1500 × 0.20 = 300 m.
- The pulse travels down and back.
- Depth = 300/2 = 150 m.
150 m below the transducer.
Look for these in your work
- Count the return leg.
- State the reference.
A 0.14 s recorded delay contains 0.02 s electronics time. Find depth below the transducer at 1500 m/s.
Give me a hint
Subtract the supplied non-travel delay first.
Compare my reasoning
- Sound time = 0.14 − 0.02 = 0.12 s.
- Total water path = 180 m.
- Depth = 90 m.
90 m below the transducer.
Look for these in your work
- Correct the interval.
- Halve only the sound path.
The corrected model depth is 90 m below a transducer 2 m below the surface. Find seabed depth below the surface.
Give me a hint
Add the stated transducer offset to its downward reflector distance.
Compare my reasoning
- Surface to transducer is 2 m.
- Transducer to seabed is 90 m.
- Surface depth = 2 + 90 = 92 m.
92 m below the surface.
Look for these in your work
- Keep both references explicit.
- Use only the supplied offset.
Common mistakes
- Count outward and return paths.
- Use the supplied water speed.
- State whether depth is below transducer or surface.
What can you explain now?
Does a slanted reflected range automatically equal vertical depth?
Compare with the explanation
No.
Additional geometry is needed to relate the slanted path to vertical distance.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain sonar and calculate distance with correct path count, medium, delay correction and reference position.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.