Question

Question: A bat emits a series of high-frequency sound pulses as it approaches a moth. The pulses are approximately \({\bf{70}}{\bf{.0}}\,{\bf{ms}}\) apart, and each is about \({\bf{3}}{\bf{.0}}\,{\bf{ms}}\) long. How far away can the moth be detected by the bat so that the echo from one pulse returns before the next pulse is emitted?

Short answer

The moth should be detected \(11.5\,{\rm{m}}\)away.

Step-by-step solution

Concept

Echo: The reflected sound heard by an observer after minimum 0.1 s of the original incident sound is known as echo.

Given Data

The chirp is \(t = 3.0\,{\rm{ms}}\) long.

The difference between one to the next pulse is \(T = 70.0\;{\rm{ms}}\).

Calculation

It is \(70.0\,{\rm{ms}}\) from the beginning of one chirp to the beginning of the next.

It is \(t' = \left( {70.0\,{\rm{ms}} - 3.0\,{\rm{ms}}} \right) = 67.0\,{\rm{ms}}\) from the end of one chirp to the beginning of the next. That means the time it takes for the pulse to travel from the bat to the moth and back is\(67.0\,{\rm{ms}}\). The sound should reach the moth and return during the\(67.0\,{\rm{ms}}\).

The distance to the moth is half the distance that the sound can travel\(67.0\,{\rm{ms}}\). The distance will be,

\(d = \frac{1}{2}vt'\)

Here,\(v = 343\;{\rm{m/s}}\)is the speed of sound in air.

Substitute the values and get,

\(\begin{array}{c}d = \left( {343\,{\rm{m/s}}} \right)\frac{1}{2}\left( {67.0\,{\rm{ms}}} \right)\\ = \left( {343\,{\rm{m/s}}} \right)\frac{1}{2}\left( {67.0 \times {{10}^{ - 3}}\,{\rm{s}}} \right)\\ \approx 11.5\;{\rm{m}}\end{array}\)

Hence, the moth should be detected\(11.5\;{\rm{m}}\)away.