Echolocation

Key Stage 4

Meaning

Echolocation is a method for finding objects using reflected sound waves.

About Echolocation

Echolocation takes place in several steps:

1. A sound is emitted.
2. The sound travels until it hits an object.
3. The sound is reflected by the object.
4. The sound travels back to the emitter.
5. The sound is detected.

The time it takes the sound to return to its source can show how far away the object is located. Since the sound has to travel to the object and back again this time is twice as long as the time taken just to travel to the object.

An object can be made 'invisible' to echolocation if it absorbs the sound instead of reflecting it.

Echolocation is used by:

• Bats - To avoid obstacles when flying and to find prey.
• Dolphins - To locate prey and the size and shape of their surroundings.
• Boats - To find shoals of fish and to find the distance to the sea bed.
• Submarines - To navigate underwater detecting the sea bed and to locate ships.

Equation

NB: You do not have to remember the equation in this form, but should be able to work it out for yourself and apply it.

Distance = (Speed) x (Time)/2

The time is divided by two because the sound wave must travel to the object and then back again to the detector.

\(x = v \frac{t}{2}\)

Where

\(v\) = The speed of sound through that medium.

\(\frac{t}{2}\) = The time taken for the sound to travel to the object or from the object back to the detector.

\(x\) = The distance between the emitter and object.

Example Calculations

A fishing boat uses sonar to detect schools of fish beneath the boat. A sonar pulse is emitted and then received 0.042 seconds later. The speed of sound in water is 1500m/s. Calculate how far beneath the boat the school of fish is located, correct to two significant figures.	A bat uses sonar to locate its prey. The bat emits an ultrasound squeak which it hears reflected off its prey 30ms later. The speed of sound in air is 340m/s. Calculate the distance to the bats prey, correct to two significant figures.
1. State the known quantities v = 1500m/s t = 0.042s	1. State the known quantities v = 340m/s t = 30ms = 30x10-3s
2. Substitute the numbers into the equation and solve. \(x = v \frac{t}{2}\) \(x = 1500 \frac{0.042}{2}\) \(x = 31.5m\) \(x \approx 32m\)	2. Substitute the numbers into the equation and solve. \(x = v \frac{t}{2}\) \(x = 340 \frac{30 \times 10^{-3}}{2}\) \(x = 5.1m\)