1.

What is beats? Give an analytical description of the phenomenon of beats. Show that the beat frequency is equal to the difference of frequencies of the component oscillation.

Answer»
  • Definition: The formation of periodic waxing and waning of sound due to interference (superposition) of two sound waves of the same amplitude but of slightly different frequencies, is called beats. 
  • One waxing and one waning which are consecutive form one beat. 
  • The time interval between two successive maxima or minima (waxings or wanings) is called the period of beats. 
  • The number of beats heard per second is called the frequency of beats.

Let us consider two waves of slightly different frequencies n1 and n2 (n1 ~ n2 < 10) having equal amplitude travelling in a medium in the same direction.  

At time t = 0, both waves travel in same phase. The equations of the two waves are  

y1 = a sin ω1t  

y1 = a sin (2πn1)t         …... (1)  

y2 = a sin ω2t  

= a sin (2πn2)t              …... (2)

When the two waves superimpose, the resultant displacement is given by  

y =  y1 + y2  

y = a sin (2πn1)t + a sin (2πn2)t        …... (3)  

Therefore  

y = 2a sin 2π(n1+ n2/2)t cos 2π(n1– n2/2)t            …... (4)  

Substitute A = 2a cos 2π(n1– n2/2)t   and n = n1 + n2/2 in equation (4)  

y = A sin 2πnt  

This represents a simple harmonic wave of frequency n = n1 + n2/2 and amplitude A which changes with time.  

(i) The resultant amplitude is maximum (i.e) ± 2a, if  

cos 2π [n1-n2/2] t = ±1  

So, 2π [n1-n2/2] t = ±mπ  

(Here, m = 0,1,2....) or (n1 – n2)t = m   

The first maximum is obtained at t1 = 0  

The second maximum is obtained at,  

 t2 = 1/n1 – n2  

The third maximum at t3 = 2/n1 – n2 and so on.  

The time interval between two successive maxima is,  

t2 – t1 =  t3 – t2 = 1/n1 – n 

Hence the number of beats produced per second is equal to the reciprocal of the time interval between two successive maxima.



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