Beats

BEATS

When two interacting waves have slightly different frequencies the resultant disturbance at any point due to the superposition periodically fluctuates causing waxing and waning in the resultant intensity. The waxing and waning in the resultant intensity of two superposed waved of slightly different frequency are known as beats.

Let the dilacement produced at a point by one wave b

y₁ = A sin(2$\pi$f₁t – $\varphi$)

And the dilacement produced at the point produced by the other wave of equal amplitude as

y₂ = A sin(2$\pi$f₂t – $\varphi$₂)

By the principle of superposition, the resultant dilacement is

Y = y₁ + y₂ = A sin(2$\pi$f₁t – $\varphi$₁) + A sin (2$\pi$f₂t – $\varphi$₂)

$y=2A\sin \left\{2\pi \left(\frac{f_1+f_2}{2}\right)t-\left(\frac{{\varphi }_1+{\varphi }_2}{2}\right)\right\}COS2\pi \left(\frac{f_1-f_2}{2}\right)t$

$Y=R\sin \left\{2\pi \left(\frac{f_1+f_2}{2}\right)t-\left(\frac{{\varphi }_1+{\varphi }_2}{2}\right)\right\}$

Where,

$R=2A\cos 2\pi \left(\frac{f_1+f_2}{2}\right)t$

The time for one beat is the time between consecutive maximum or minima.

First maxima would occur when

$\cos 2\pi \left(\frac{f_1+f_2}{2}\right)t=+1$

Then $2\pi \left(\frac{f_1-f_2}{2}\right)t=0$

or $t=\frac{1}{f_1-f_2}$

The time for one beat = $\left(\frac{1}{f_1-f_2}-0\right)=\frac{1}{f_1-f_2}$

Similarly it may also be shown that time between two consecutive minima is $\frac{1}{f_1-f_2}$.

Hence, frequency of beat i.e. number of beats in one second or

Beat frequency = f₁ $\sim$ f₂.

Illustration 1: A column of air at 51⁰C and a tuning fork produce 4 beats per sec when sounded together. As the temperature of the air column is gradually decreased to 16⁰C, they again produce 1 beat/s. Find the frequency of the tuning fork. It is found that they again produce 1 beat/s if temperature is further decreased explain.

Solution: Let n = frequency of the tuning fork:

$\Rightarrow$ $\frac{n+4}{n+1}=\sqrt{\frac{273+51}{273+16}}$

The number of beats per sec decreases

$\Rightarrow$ n = 50 Hz

as the frequency of the air column decreases. They can again produce 1 beat/s when

Hence the frequency of air column is initially air column frequency reduced to [n – 1]

(n + 4) at 51⁰C. At 16^oC, the frequency of air at a further low temperature T.

column can be (n + 1).

$\Rightarrow$ n + 4 $\alpha$ c₅₁

$\Rightarrow$ n + 4 $\alpha$ c₁₆

Where c is the speed of sound

$\frac{n+1}{n-1}$ = $\sqrt{\frac{273+16}{T}}$