A) 0
B) 1
C) 2
D) 3
Correct Answer: B
Solution :
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 \[y=a\sin 2\pi (n-1)t,\] \[y=a\sin 2\pi nt\] and \[{{y}_{3}}=a\sin 2\pi (n+1)t\] Resultant displacement due to all three waves is \[y={{y}_{1}}+{{y}_{2}}+{{y}_{3}}\] \[=a\sin 2\pi nt+a\,[\sin 2\pi (n-1)t+\sin 2\pi (n+1)t]\] \[=a\sin 2\pi nt+a\,[2\sin 2\pi nt\cos 2\pi t]\] \[\left[ \text{Using }\sin C+\sin D=2\sin \frac{C+C}{2}\cos \frac{C-D}{2} \right]\] Þ \[y=a\,(1+\cos 2\pi t)\sin 2\pi nt\] This is the resultant wave having amplitude \[=(1+\cos 2\pi t)\] For maximum amplitude cos 2pt = 1 Þ 2pt = 2mp where m = 0, 1, 2, 3, ... Þ t = 0, 1, 2, 3 ... Hence time interval between two successive maximum is 1 sec. So beat frequency = 1 Also for minimum amplitude (2cos 2pt) = 0 Þ \[\cos 2\pi t=-\frac{1}{2}\] Þ \[2\pi t=2m\pi +\frac{2\pi }{3}\] Þ \[t=+\frac{1}{3}\] Þ \[t=\frac{1}{3},\frac{4}{3},\frac{7}{3},\frac{10}{3},....\] (for m = 0, 1, 2, ..) Hence time interval between two successive minima is 1 sec so, number of beats per second = 1 Note : PET/PMT Aspirants can remember result only.
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