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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?

Nikhila Thakur , 9 Years ago
Grade 12th pass
anser 2 Answers
12Myra Kothari
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
 
 
 
Last Activity: 7 Years ago
Rishi Sharma
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
Last Activity: 5 Years ago
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