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Beats - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Beats is considered one of the most asked concept.

  • 28 Questions around this concept.

Solve by difficulty

A tuning fork arrangement (pair) produces 4 beats/sec with one fork of frequency 288 cps. A little wax is placed on the unknown fork and it then produces 2 beats/sec. The frequency of the unknown fork is _____ cps.

We have 3 forks A, B, and C respectively. And their frequency is in AP. If A & B forms a tuning fork & A&C forms another tuning fork pair  Then what is the ratio of beat frequency tuning fork pair 

What is the necessary condition for Beat phenomena when two sound waves A & B have the same amplitude and travelling in a medium in the same direction 

Frequency of tuning forks are 320 Hz and 325 Hz. If they are sounded together, the beat period is

To demonstrate the phenomenon of beats we need

Concepts Covered - 1

Beats

Beats -

When any two sound waves of slightly different frequencies, travelling along the same direction in a medium and superimpose on each other then the intensity of the resultant sound at a particular position rises and falls regularly with time. This phenomenon of regular variation in intensity of sound with time at a particular position is called beats. 

If we struck two tuning forks of slightly different frequencies, one hears a sound of periodically varying amplitude. This phenomenon is called beating.

Beat frequency, equals the difference in frequency between the two sources which we will see below.

Let us consider two sound waves travelling through a medium having equal amplitude with slightly different frequencies $f_1$ and $f_2$. We use equations similar to equation $y=A \sin (k x-\omega t)$ to represent the wave functions for these two waves at a point such that $k x=\pi / 2$.

$
\begin{aligned}
& y_1=A \sin \left(\frac{\pi}{2}-\omega_2 t\right)=A \cos \left(2 \pi f_1 t\right) \\
& y_2=A \sin \left(\frac{\pi}{2}-\omega_2 t\right)=A \cos \left(2 \pi f_2 t\right)
\end{aligned}
$


By using superposition principle -

$
y=y_1+y_2=A\left(\cos 2 \pi f_1+\cos 2 \pi f_2 t\right)
$


We can also write the above equation by using trigonometric identity as -

$
y=\left[2 A \cos 2 \pi\left(\frac{f_1-f_2}{2}\right) t\right] \cos 2 \pi\left(\frac{f_1+f_2}{2}\right) t
$
 

The graph is like this - 

                                                      

Graphs of the individual waves and the resultant wave are shown in the figure. We can see that the resultant wave has effective frequency equal to average frequency $\frac{f_1+f_2}{2}$. From the figure, we can see that this wave is multiplied by the envelope whose equation is given as -

$
y_{\text {envelope }}=2 A \cos 2 \pi\left(\frac{f_1-f_2}{2}\right) t
$


A maximum in the amplitude of the resultant sound wave is detected whenever

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


Hence, there are two maxima in each period of the envelope wave. Because the amplitude varies with frequency as $\frac{\left(f_1-f_2\right)}{2}$ the beat frequency is two times of this value and given by -

$
f_{\text {beat }}=\left|f_1-f_2\right|
$
 

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Beats

Physics Part II Textbook for Class XI

Page No. : 382

Line : 29

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