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Chapter 6: Q90GP (page 138)

Question: Two loudspeakers face each other at opposite ends of a long corridor. They are connected to the same source which produces a pure tone of\({\bf{282}}\,{\bf{Hz}}\). A person walks from one speaker toward the other at a speed of \({\bf{1}}{\bf{.6}}\,{\bf{m/s}}\).What 鈥渂eat鈥� frequency does the person hear?

Short Answer

Expert verified

The beat frequency heard is \({\rm{2}}{\rm{.6}}\,{\rm{Hz}}\).

Step by step solution

01

Concept

Beat: The difference in frequency of the two superposed sound wave is known as beat.

02

Given Data

The frequency of the source is\(f = 282\;{\rm{Hz}}\).

The speed of the person towards the speaker is \({v_{walk}} = 1.6\;{\rm{m/s}}\).

03

Calculation

The apparent frequency when the person walks towards the speaker is,

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

The apparent frequency when the person walks away from the speaker is,

The beat frequency is,

Substitute the values and get,

Hence, the beat frequency heard is \({\rm{2}}{\rm{.6}}\,{\rm{Hz}}\).

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