Q1.
An experiment is being set up that requires echoes to arrive at least 0.2 s after the emission of sound. What minimum distance should a reflecting surface be placed at? Assume the speed of sound to be 343 m s–1.
Answer
The reflecting surface must be at least 34.3 m away.
The sound has to make a round trip: from the source to the wall and back to the listener. So the total path is twice the distance we are asked for.
total distance travelled = speed × time
= 343 m s⁻¹ × 0.2 s
= 68.6 m
This is the to-and-fro distance, so
distance to the reflecting surface = 68.6 m ÷ 2
= 34.3 m
= 343 m s⁻¹ × 0.2 s
= 68.6 m
This is the to-and-fro distance, so
distance to the reflecting surface = 68.6 m ÷ 2
= 34.3 m
Written as a single formula, d = v t / 2 = (343 m s⁻¹ × 0.2 s) / 2 = 34.3 m. Any wall closer than this returns the echo too soon for the experiment.
Why the factor of 2 is not optional: forgetting it doubles the answer to 68.6 m. Every echo, sonar and ultrasonic-sensor calculation in this chapter has the same structure — the measured time is always the round-trip time, so always halve either the time or the distance, never neither and never both.
Tip: compare this with the book's own case. The minimum distance for hearing any echo at all is set by the ear's 0.1 s limit: d = (340 m s⁻¹ × 0.1 s) / 2 = 17 m. This experiment simply demands twice the delay, so it needs roughly twice the distance.