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
Text Solution
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Let n – 1 (= 400), n (= 401) and n + 1 (= 402) be the frequencies of the three waves. If a be the amplitude of each then
and 
Resultant displacement due to all three waves is 



⇒ ⇒ 
This is the resultant wave having amplitude 
For maximum amplitude cos 2 π π t = 1 ⇒ ⇒ 2 π π t = 2m π π where m = 0, 1, 2, 3, ...
⇒ ⇒ t = 0, 1, 2, 3 ...
Hence time interval between two successive maximum is 1 sec. So beat frequency = 1
Also for minimum amplitude (2cos 2 π π t) = 0

Hence time interval between two successive minima is 1 sec so, number of beats per second = 1
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