Question

What is beats.prove that the number of beats per second is equal to the difference between the frequencies​

Answer 1

Beats

When two sound waves of equal amplitude but of nearly same frequencies (frequencies having a difference less than 10 Hz) superimpose each other, the amplitude of the new sound produced changes from maximum to minimum and the process continues, called waxing and waning, respectively. One waxing and one waning together constitute a beat.

The no. of beats produced per second is called beat frequency and is given by,

$\displaystyle\longrightarrow\sf {\nu_{beat}=|\nu_1-\nu_2|\quad\quad\dots (1)}$

Proof for the expression of beat frequency:-

Let a beat be produced by the two individual progressive sound waves,

$\displaystyle\longrightarrow\sf {y_1=A\sin(2\pi\nu_1t)}$

and,

$\displaystyle\longrightarrow\sf {y_2=A\sin(2\pi\nu_2t)}$

Then the beat produced is given by,

$\displaystyle\longrightarrow\sf {y_1+y_2=A\sin(2\pi\nu_1t)+A\sin (2\pi\nu_2t)}$

$\displaystyle\longrightarrow\sf {y_1+y_2=A\left [\sin(2\pi\nu_1t)+\sin (2\pi\nu_2t)\right]}$

But,

$\displaystyle\longrightarrow\sf {\sin c+\sin d=2\sin\left (\dfrac {c+d}{2}\right)\cos\left (\dfrac {c-d}{2}\right)}$

Then,

$\displaystyle\longrightarrow\sf {y_1+y_2=2A\sin\left(\dfrac {2\pi\nu_1t+2\pi\nu_2t}{2}\right)\cos\left(\dfrac {2\pi\nu_1t-2\pi\nu_2t}{2}\right)}$

$\displaystyle\longrightarrow\sf {y_1+y_2=2A\sin\left(\pi(\nu_1+\nu_2)t\right)\cos\left(\pi(\nu_1-\nu_2)t\right)}$

Or,

$\displaystyle\longrightarrow\sf {y_1+y_2=2A\cos\left(\pi(\nu_1-\nu_2)t\right)\sin\left(2\pi\left(\dfrac {\nu_1+\nu_2}{2}\right)t\right)}$

Let $\displaystyle\sf {y_1+y_2=y',\ 2A\cos\left(\pi(\nu_1-\nu_2)t\right)=A'}$ and $\displaystyle\sf {\dfrac {\nu_1+\nu_2}{2}=\nu'.}$ Then,

$\displaystyle\longrightarrow\sf {y'=A'\sin (2\pi\nu't)}$

Here we see that the amplitude of the new sound produced is $\displaystyle\sf {A'=2A\cos\left(\pi(\nu_1-\nu_2)t\right)}$ which increases or decreases by time.

We know that waxing occurs at maximum amplitude, i.e., at $\displaystyle\sf {A'=\pm2A.}$

Thus for maximum amplitude,

$\displaystyle\longrightarrow\sf {\cos\left(\pi(\nu_1-\nu_2)t\right)=\pm1}$

$\displaystyle\longrightarrow\sf {\pi(\nu_1-\nu_2)t=n\pi}$

$\displaystyle\longrightarrow\sf {\nu_1-\nu_2=\dfrac {n}{t}\quad\quad\dots (2)}$

where $\displaystyle\sf {n}$ is probably the no. of beats produced under the time $\displaystyle\sf {t.}$

So (2) actually shows that the no. of beats produced per second is equal to the difference of frequencies of the individual sound waves.

Combining (1) and (2), we get,

$\displaystyle\longrightarrow\sf {\nu_{beat}=\dfrac {\sf{No.\ of\ beats\ produced}}{\sf{Time}}=|\nu_1-\nu_2|}$

Hence the Proof!

Answer 2

Answer:

prove that the number of beats heard per second is equal to the difference in frequencies of two sound sources