wave_mechanics critical_thinking Question 46
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
[MNR 1980; J & K CET 2005]
Options:
A) 0
B) 1
C) 2
D) 3
Show Answer
Answer:
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] $
$ [ \text{Using }\sin C+\sin D=2\sin \frac{C+C}{2}\cos \frac{C-D}{2} ] $
Therefore $ 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
Therefore 2pt = 2mp where m = 0, 1, 2, 3, …
Therefore 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
Therefore $ \cos 2\pi t=-\frac{1}{2} $
Therefore $ 2\pi t=2m\pi +\frac{2\pi }{3} $
Therefore $ t=+\frac{1}{3} $
Therefore $ 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.
BETA
