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The frequency of note emitted by a source changes by 20% as it approaches observer.As it recedes away from him the apparent frequncy will be?

prema , 6 Years ago
Grade 12th pass
anser 2 Answers
prema

Last Activity: 6 Years ago

mein ne is question ko post kiya tha.par abtak kuch answer nahi mila.jab meine askiitans mein login kiya tha tab yah sochlar kiya tha ki yahanke teachers mere is question par gaur karrnge aur mujhe answe
r denge.lekin abtak meine jitne hi questions
dale hai muje koi javab nahi mila hai.yeh
mein ne nahi socha tha

Arun

Last Activity: 6 Years ago

Dear student
 
 
n(120/100) = (v / v – vs) *n
1.2 V – 1.2 Vs = V
0.2 V = 1.2 Vs
Vs = V/6
when it recedes away
(100 – x / 100) * n = (V / V + Vs)*n
1 – x /100 = 6/7
x = 14.3 %
hence 14.3% different from actual frequency
 
 
Regards
Arun

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