A) 1
B) 2
C) 3
D) 0
Correct Answer: A
Solution :
The resultant displacement of a particle by these three waves, \[y=a\sin 2\pi 400t+a\sin 2\pi 401t+a\sin 2\pi 402t\] \[y=a[\sin 2\pi 400t+a\sin 2\pi 402t+a\sin 2\pi 401t\] \[=2a\sin 2\pi 401t\cos 2\pi t+a\sin 2\pi 401t\] \[=a(1+2\cos 2\pi t)\sin 2\pi 401t\] So the resultant amplitude is\[a(1+2\cos \pi t)\] Resultant amplitude is maximum when \[\cos 2\pi t=1\] or \[2\pi t=0,\,\,2\pi ,\,\,4\pi \] \[\therefore \] \[t=0,\,\,1,\,\,2,\,\,3,\,\,....\] Thus, the time interval between two successive maxima is\[1\,\,\sec \]. Similarly resultant amplitude is minimum when \[t=\frac{1}{2},\,\,\frac{3}{2},\,\,\frac{5}{2},\,\,...\] So, beat frequency is \[1\] per sec
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