Question

Show analytically the beat frequency is equal to the difference between the reciprocal of the period of the two interfering notes

Answer 1

Explanation:

When two waves of nearly equal frequencies travelling in a medium along the same direction superimpose upon each other, beats are produced. The amptitude of the resultant sound at a point rises and falls regularly.

The intensity of the resultant sound at a point rises and falls regularly with time. When the intensity rises to maximum we call it as waxing of sound, when it falls to minimum we call it as waning of sound.

The phenomenon of waxing and waning of sound due to interference of two sound waves of nearly equal frequencies are called beats. The number of beats produced per second is called beat frequency, which is equal to the difference in frequencies of two waves.

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Answer 2

The beat frequency is equal to the difference between the reciprocal of the period of the two interfering notes.

Explanation:

Beat frequency :

The beat frequency is equal to the difference between the reciprocal of the period of the two interfering notes.

When time lapsed between two successive sound it is called beat period.

Reciprocal of beat period is called beaqt frequency.

Let two waves be represented by

$y_{1}=a\sin\omega_{1}t$

$y_{1}=a\sin2\pi v_{1}t$....(I)

$y_{2}=a\sin2\pi v_{2}t$....(II)

If y is the resultant of displacement due to superposition then

$y=y_{1}+y_{2}$

$y=a\sin2\pi v_{1}t+a\sin2\pi v_{1}t$

$y=2a\sin\pi(v_{1}+v_{2})t\cos\pi(v_{1}+v_{2})t$

$y=(2a\cos\pi(v_{1}-v_{2})t)\sin\pi(v_{1}+v_{2})t$

$y=A\sin\pi (v_{1}+v_{2})t$

Here, $A=2a\cos\pi(v_{1}-v_{2})t$

A is the amplitude of the resultant wave.

Since intensity is directly proportional to square of amplitude.

Therefore, intensity is maxiumu where A is maximum.

$\cos\pi(v_{1}-v_{2})t=\pm 1$

$\pi(v_{1}-v_{2})=n\pi$

$t=\dfrac{n}{v_{1}-v_{2}}$

Intensity will be maximum

At $t = 0, \dfrac{1}{v_{1}-v_{2}}, \dfrac{2}{v_{1}-v_{2}}......$

So, The time interval between two successive sound is

$T=\dfrac{1}{v_{1}-v_{2}}$

We need to calculate the beat frequency

Using formuka of frequency

$b_{f}=\dfrac{1}{T}$

Put the value into the formuka

$b_{f}=v_{1}-v_{2}$

Hence, The beat frequency is equal to the difference between the reciprocal of the period of the two interfering notes.

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Topic : time period

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