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Calculate the frequency of beats produced in the air when two sources of sound are activated, one...

Question:

Calculate the frequency of beats produced in the air when two sources of sound are activated, one emitting a wavelength of 32 cm and the other of 32.2 cm. The speed of sound in air is 350 m s-1.

Beats

When two waves of nearly the same amplitude and slightly different frequencies act simultaneously then a receiver at a fixed location will record an oscillation in the amplitude of the received resultant wave. The frequency of these oscillations is called the beat frequency. Beats are nothing but interference in the time domain. If the frequency of one wave is {eq}\displaystyle {\nu_1} {/eq} and that of the other is {eq}\displaystyle {\nu_2} {/eq} then the beat frequency is {eq}\displaystyle {\nu=\nu_1-\nu_2} {/eq}.

Answer and Explanation:

The propagation speed of a wave {eq}\displaystyle {v} {/eq}, the frequency {eq}\displaystyle {\nu} {/eq} and the wavelength {eq}\displaystyle {\lambda} {/eq} are related according to ,

{eq}\displaystyle {v=\nu \lambda} {/eq}.

Here it is given that the speed of sound in air is 350 m/s. The wavelength of the first wave is 0.32 m. therefore the frequency is,

{eq}\displaystyle {\nu_1=\frac{350}{0.32}=1093.75\ Hz} {/eq}.

The second wave has a wavelength 0.322 m. Therefore,

{eq}\displaystyle {\nu_2=\frac{350}{0.322}=1086.95\ Hz} {/eq}

Therefore the beat frquency is,

{eq}\displaystyle { \nu=1093.75-1086.95=6.8\ Hz} {/eq}.


Learn more about this topic:

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The Resultant Amplitude of Two Superposed Waves

from MEGA Physics: Practice & Study Guide

Chapter 15 / Lesson 10
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