Question

show analytically that beat frequency is equal to the difference between the reciprocal of the period of the two interfering notes​

Answer 1

$f_{b}=f_{2}-f_{1}$

This is the beat frequency formula of the two waves.

Explanation:

We know that,

The formula of the frequency is reciprocal of the period.

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

Where f= frequency

T = period

We know that,

The beat frequency is defined as,

$f_{b}=(f_{1}-f_{2})$

Where $f_{1} and\ f_{2}$ are frequencies of the two interfering notes​.

We need to show the beat frequency is equal to the difference between the reciprocal of the period of the two interfering notes​

Let us consider that one source has a shorter time period as $T_{s}$ and high frequency as $f_{2}$  whereas other source time periods and frequency are $T_{l}$ and $f_{1}$.

We know that,

The change time period is

Δ$T$=$T_{l}-T_{s}$

We need to write an equation for short period and long period

We need to write an equation for short period and long period

$n\times \Delta T=T_{l}$   -------(a)

$T_{b}=n\times T_{s}$   -------(b)

Put the value of n in equation (b) from equation (a)

$T_{b}=\dfrac{T_{l}}{\Delta T}\times T_{s}$

Put the value of Δ$T$

$T_{b}=\dfrac{T_{l}}{T_{l}-T_{s}}\times T_{s}$

$T_{b}=\dfrac{T_{l}T_{s}}{T_{l}-T_{s}}$

Now, dividing top and bottom by the product of $T_{l}T_{s}$

$T_{b}=\dfrac{1}{\dfrac{1}{T_{s}}-\dfrac{1}{T_{l}}}$

Taking reciprocal both side,

$\dfrac{1}{T_{b}}=\dfrac{1}{T_{s}}-\dfrac{1}{T_{l}}$

Using relation of frequency

$f_{b}=f_{2}-f_{1}$

This is the beat frequency formula of the two waves.