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Nilanjan Bhowmick AIR 3, CSIR NET (Earth Science)
- IIT JAM
- Physics (PH)
(b) the given equation can be rewritten as [d+(g/eje = 0, where d d/dt the auxiliary equation is m? + (g/e) = 0 :om. m2- (g/e) or m = t v(g/e)ji. the solution o
(B) The given equation can be rewritten as [D+(g/eje = 0, where D d/dt The auxiliary equation is m? + (g/e) = 0 :om. m2- (g/e) or m = t v(g/e)ji. The solution of the given, differential equation is e C, cos {V(g/) t+C) where C,, C, are arbitrary constants ) and de/dt=- V(g/e)C, sin fV(g/e) +C} where t= 0, 0 = 0, and de/dt = 0 (given so from () and (i) we get 6=C, cos C, and 0 = -V(g/e)C, sin C From these we have sin C, = 0 and C, # 0 and so C, = From (), the required solution is 0 , cos {(V(g/e) t)}
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Ravi bhushan![best-answer]()
Here is the detailed solution.
please explain why u write theta =C1 cos {√g/l+ C2 }
for imaginary part what is c.f formula
For imaginary part CF formula is Yçf = A CosX +B SinX
But in your question, Only Cosine part is given.