Use the approximate speed of sound in air
Recall the 330–350 m/s range and distinguish propagation speed from vibration frequency and a universal constant.
Section P3.4 · Sound
Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.
Before you begin
- Use speed = distance/time.
- Distinguish sound and electromagnetic waves.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which unit describes propagation speed?
What you will learn
- State the approximate 330–350 m/s air sound speed.
- Calculate direct-path time or distance using a stated speed.
Use an approximate air value
The speed of sound in air is approximately 330–350 m/s under the ordinary conditions represented in this syllabus. A problem may supply 340 m/s as a working value. Temperature and medium conditions matter, so this is not the same kind of vacuum constant as electromagnetic speed.
A 68 m direct sound path at 340 m/s takes 68/340 = 0.20 s. This uses the one-way path from source to receiver. An echo travels outward and back, which needs a different path count.
Distinguish frequency and travel
A source frequency measures vibrations per second; propagation speed measures metres per second. Changing the pitch in the same ordinary air conditions does not require a different propagation speed in the school model. A 500 Hz and 1000 Hz sound can use the same stated air speed.
An observer seeing a flash and hearing a sound later can use the delay as approximately the sound travel time when light travel is negligible over the stated distance. Source synchronisation, reaction timing and the actual path remain measurement considerations.
Sources for this lesson
Pause and explain
At 340 m/s, how long does a direct 68 m sound path take?
Worked example
Find the direct sound travel time for 68 m at 340 m/s.
Show the worked solution
- Use the one-way path d = 68 m.
- t = d/v = 68/340.
- t = 0.20 s.
Answer 0.20 s.
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.
State the approximate speed range of sound in air.
Give me a hint
Include a speed unit rather than a frequency unit.
Compare my reasoning
- The lower approximate reference is 330 m/s.
- The upper is 350 m/s.
- These describe ordinary air conditions.
Approximately 330–350 m/s.
Look for these in your work
- Give the range.
- Retain the approximation.
Calculate the direct distance sound travels in 0.50 s at 340 m/s.
Give me a hint
Use distance = speed × time for one direct path.
Compare my reasoning
- v = 340 m/s.
- d = vt = 340 × 0.50.
- d = 170 m.
170 m.
Look for these in your work
- Use the stated speed.
- Give a distance unit.
A 500 Hz sound and a 1000 Hz sound travel in the same ordinary air model at 340 m/s. Must the higher-pitch sound arrive sooner over 68 m?
Give me a hint
Compare path and speed, not frequency alone.
Compare my reasoning
- Both have the same stated propagation speed.
- Both use the same 68 m path.
- Both travel times are 0.20 s.
No; the stated equal-speed model gives equal travel times.
Look for these in your work
- Use the model's condition.
- Separate pitch and speed.
Common mistakes
- Air sound speed is approximate.
- Frequency and speed have different units.
- Do not halve a direct one-way path.
What can you explain now?
What approximate air speed range should you recall?
Compare with the explanation
330–350 m/s.
Use a stated working value when a calculation provides one.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Recall air sound speed, calculate direct-path travel and explain why frequency is a separate quantity.
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.