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Two stationary sources each emitting waves of wave length λ. An observer moves from one source to other with velocity u. Then number of beats heared by him is?

Two stationary sources each emitting waves of wave length λ. An observer moves from one source to other with velocity u. Then number of beats heared by him is?

Grade:12th pass

2 Answers

12Myra Kothari
17 Points
6 years ago
f1= (v-u​)f/v
f2= (v+u)f/v
Now beat frequency = f2-f1 = \frac{(v+u)f}{v} - \frac{(v-u)f}{v} = \frac{2uf}{v}= \frac{2u}{l}
here l = wavelength 
therefore answer = 1) 2u/lambda
 
 
 
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

For first source 645-163_01.PNG
for IInd source645-1353_001.PNG
Beatfreq. = ∣n1​−n2​∣
= n + nu/v – n + nu/v
= 2nu/v​ = 2u/​λ [∵v = nλ ∵1/λ ​= n/v​ ]

Thanks and Regards

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