Question

hey please help......
prove that no. of beats per second is equal to difference BTW frequencies of 2 superimposing waves........



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Answer 1

When two waves of nearly equal frequencies travelling in a medium along the same direction superimpose upon each other, beats are produced. The amplitude of the resultant sound at a point rises and falls regularly.

The intensity of the resultant sound at a point rises and falls regularly with time. When the intensity rises to maximum we call it as waxing of sound, when it falls to minimum we call it as waning of sound.

The phenomenon of waxing and waning of sound due to interference of two sound waves of nearly equal frequencies are called beats. The number of beats produced per second is called beat frequency, which is equal to the difference in frequencies of two waves.

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 – n2

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

Beat Period

(ii) The resultant amplitude is minimum (i.e) equal to zero, if

cos 2π [n1-n2/2] t = 0

(i.e) 2π [n1-n2/2] t = π/2 + mπ = (2m+1)π/2  

Or, [n1-n2/2] t = (2m+1)/2  

Where m = 0, 1, 2...

The first minimum is obtained at,

t1' = 1/2(n1 – n2)

The second minimum is obtained at,

t2' = 3/2(n1 – n2)

The third minimum is obtained at,

t3' = 5/2(n1 – n2) and so on

Time interval between two successive mimima is

t2' –  t1' = t3' – t2' = 1/n1 – n2

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

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