Consider two waves of slightly different frequencies `v_(1)` and `v_(2)` having same amplitude A. The displacements of two wave are represented by the equations:
`y_(1) = A sin omega_(1)t = A sin 2 pi v_(1)t`
and `Y_(2) = A sinn omega_(2)t = A sin 2pi v_(2)t`
As per superposition principle, the resultant displacement is gfven by:
`Y = Y_(1) + Y_(2) = A sin 2pi v_(1)t + A sin 2piv_(2)t`
or `Y= A[sin2piv_(1)t + sin 2piv_(2)t]`
But `sinC + sin D = 2 sin (C+D)/2 cos (C-D)/2`
or, `Y = A [2sin (2pi_(1)t + 2piv_(2)t)/2 cos (2piv_(1)t - 2piv_(2)t)/2]`
or, `Y = [2A cos 2pi (v_(1)-v_(2))/2] sin 2pi ((v_(1) -v_(2)))/2] sin 2pi((v_(1)+v_(2)t))/2`
`Y =A. sin 2pi((v_(1) + v_(2)t))/2`.........(i)
Equation (i) represents a simple harmonic wave of frequency and amplitude.
`A. =2A cos 2pi ((v_(1)-v_(2)t))/2`.....(ii)
For maximum amplitude (or intensity): The amplitude of the resultant wave will be maximum, if
`cos 2pi ((v_(1)-v_(2)t)/2) =+1` or, `pi(v_(1)-v_(2))t = npi`, where n =0,1,2,3.........
or `t = n/(v_(1)-v_(2))`
Maximum amplitudes occur at time `t = 1/(v_(1)-v_(2)), 2/(v_(1)-v_(2))`,........
`therefore` Time interval between successive maxima,
`T = 2/(v_(1)-v_(2)) -1/(v_(1)-v_(2)) = 1/(v_(1)-v_(2))`
`therefore` Frequency `(v_(1) + v_(2))/2` of maximum intensity, `v = 1/2 = v_(1) - v_(2)`
For minimum amplitude (or intensity):
The amplitude of the resultant wave will be minimum, if `cos 2pi ((v_(1)-v_(2))t)/2 =0`
or, `pi (v_(1)-v_(2))t`= where n =0,1,2,3,......... or, `t = (2n+1)/(2(v_(1)-v_(2))`
Minimum amplitudes occur at times `t =1/(2(v_(1)-v_(2))) , 3/(2(v_(1)-v_(2))), 5/(2(v_(1)-v_(2))`,.........
`therefore` Frequency of minimum intensity, `v = 1/T = v_(1) - v_(2)`
.Hence frequencies of maxima and minima are equal. This is known as frequency of beat, i.e. frequency of beats, `v = v_(1) - v_(2)`
