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1. A submarine can use sonar (sound traveling through water) to determine its di

ID: 1389763 • Letter: 1

Question

1. A submarine can use sonar (sound traveling through water) to determine its distance from other objects. The time between the emission of a sound pulse (a "ping") and the detection of its echo can be used to determine such distances. Alternatively, by measuring the time between successive echo receptions of a regularly timed set of pings, the submarine's speed may be determined by comparing the time between pings. Assume you are the sonar operator in a submarine traveling at a constant velocity underwater. Your boat is in the eastern Mediterranean Sea, where the speed of sound is known to be 1522 m/s. If you sent out pings every 4.6 s, and your apparatus receives echoes reflected from an undersea cliff every 4.58 s, how fast is your submarine approaching the cliff?

2.

Bats use echolocation to determine their distance from objects they cannot easily see in the dark. The time between the emission of a high-frequency sound pulse (a click) and the detection of its echo is used to determine such distances. A bat, flying at a constant speed of 19.3 m/s in a straight line toward a vertical cave wall, makes a single clicking noise and hears the echo 0.18 s later. Assuming that she continued flying at her original speed, how close was she to the wall when she received the echo? (assume the speed of sound is 343 m/s)

Explanation / Answer

Divide by 2 because it is going and bouncing back

(4.6 - 4.56) / 2 * 1522 = 15.22 m/sec
15.22 * 3600 / 1000 = 54.792 kph

Maybe it is a jet ski, not a sub. This is faster than an atomic submarine. There must be a strong current pushing it along

54.792 * 0.539957 = 29.585 knots

2)

When you position the origin of the click, position Listening to reflection,

d = speed * time = 19.1 * 0.17 = 3.247 = 3.25m

Distance to travel is click,

d = speed * time = 343 * 0.17 = 58.31m

From the origin, the distance to the wall,

d = (58.31 +3.247) / 2 = 30.78m

From the current position, the distance to the wall,
d = 30.78-3.25 = 27.53 m