Question

prove that the number of beats produced in per second is equal to the difference between the frequencies of two superimposing waves​

Answer 1

Answer: Beat frequency is equal to the difference between the frequencies of two superimposing waves​.

Explanation:

Let the two waves have frequency f₁ and f₂

Wave 1 : Asin(2πf₁t)

Wave 2 : Asin(2πf₂t)

On superimposing the waves:

Wave 1 + Wave 2 :  Asin(2πf₁t) + Asin(2πf₂t)

Wave 1 + Wave 2 : $2Asin(\pi (f_1 + f_2)t)cos(\pi (f_1 - f_2)t)$

Resulting wave has 2 sinusoidal components.

One has frequency (f₁ - f₂) and other has frequency (f₁ + f₂).

As the frequency (f₁ - f₂) is smaller frequency , the frequency of the superimposed wave will be (f₁ - f₂).

Beat frequency = Number of maxima per second = Number of minima per second = Frequency of superimposed wave =  | f₁ - f₂ |

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