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Beats - Practice Questions & MCQ

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

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  • Beats is considered one of the most asked concept.

  • 19 Questions around this concept.

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Two identical strings are stretched at tensions \mathrm{T_A \: and \: T_B}. A turning fork is used to set them in vibration. A vibrates in its fundamental mode and \mathrm{B} in its second harmonic mode then -
 

The first overtone frequency of a closed organ pipe is equal to the fine overtone frequency of an open organ pipe. Further \mathrm{ n^{t h}} harmonic of a closed organ pipe is also equal to the \mathrm{m^{\text {th }}} harmonic of an open pipe, where \mathrm{n \: \: and \: \: m} are-
 

Two identical string instruments have frequency of 100 \mathrm{~Hz}. If the tension in one of them increases 4 % and they are sounded together, then the number of beats - in 1\: \mathrm{sec} are -
 

The frequency of tuning forks A and B are respectively 3% more and 2% less than the frequency of tuning fork C. When A and B are simultaneously excited, 5 beats per second are produced. Then the frequency of the tuning fork A (in Hz) is -

Two closed pipes produce 10 beats per second when emitting their fundamental nodes. If their lengths are in the ratio of 25:26, then the ratio of fundamental frequency-

An organ pipe, open from both ends produces 5 beats per second when vibrating with a source of frequency 200 Hz in its fundamental mode. The second harmonic of the same pipe produces 10 beats per second with a source of frequency 420 Hz. The fundamental frequency of pipe is:-

A source of sound S of frequency 500 Hz situated between a stationary observer O and a wall, moves towards the wall with a speed of 2 m/s. If the velocity of sound is 332 m/s, then the number of beats per second heard by the  observer is (approximately)-

 

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When a stretched wire and a tuning fork are sounded together Y beats Per second are produced When the length of the wire is  90  cm or 100 cm frequency of the fork is 195 Hz, Then Y =?
 

Two turning forks of frequencies f1 and f2 produce 'f beats' per second. If f2 and f are known,fmay be given by :

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When a stretched wire and a tuning fork are sounded together, y beats per second are produced when the length of the wire is 85 cm or 90 cm, frequency of the fork is 180 Hz. Find the value of y.

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{array}{l}{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{array}

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

 \begin{array}{l}{\text { A maximum in the amplitude of the resultant sound wave is }} {\text { detected whenever }} \\ \\ {\qquad \cos 2 \pi\left(\frac{f_{1}-f_{2}}{2}\right) t=\pm 1}\end{array}

Hence, there are two maxima in each period of the envelope wave. Because the amplitude varies with frequency as \frac{(f_{1}-f_{2} ) }{ 2} the beat frequency is two times of this value and given by  - 

                                                                                     f_{\mathrm{beat}}= |f_{1}-f_{2} |

                                                               

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Beats

Physics Part II Textbook for Class XI

Page No. : 382

Line : 29

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