Question

There are three sources of sound of equal intensity with frequencies 400, 401 and 402 vib/sec. The number of beats heard per second is​

Answer 1

Explanation:

Given  There are three sources of sound of equal intensity with frequencies 400, 401 and 402 vib/sec. The number of beats heard per second is​

• So we have ν – 1 = 400 so y = a sin 2 π (n – 1) t
•          So   ν = 401 so y = a sin nt
•        So ν + 1 = 402 so y = a sin 2 π (n + 1) t
•   Now y = y1 + y2 + y3
•              = a sin 2 π (n – 1) t + [a sin nt + y = a sin 2 π (n + 1) t]
• So we have sin C + sin D = 2 sin C + D/2 cos C – D/2
•               So y = a (1 + cos 2πt) sin 2πnt
• For maximum amplitude 1 + cos 2 πt = 2
• So cos 2πt = 1
• So 2πt = 2mπ
• Here m = 0,1,2,3……..
• So t = 0,1,2,3………
• Therefore time between two successive maxima is 1 sec
• Now for minimum amplitude we have
• Cos 2πt = - 1/2  
• So 2πt = 2mπ + 2π/3
• So t = m + 1/3
• So m = 0,1,2,3…..
• So t = 1/3, 4/3, 7/3, 10/3….
• So time between two successive minima is 1 sec.
• Therefore beats per second will be = 1

Reference link will be

https://brainly.in/question/14544337