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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Feb 2008 18:11:50 IST
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S.H.M-Q.30??(H.C.VERMA)
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7 Feb 2008 18:03:47 IST
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no ans?????
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7 Feb 2008 19:09:49 IST
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post quest?
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A man without faith is like a bird without wings
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7 Feb 2008 19:31:30 IST
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do post q
every one do not have HCV
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b)
from conservation of momentum-
mx1=Mx2 --------------1
m(v1-v2)=M(v2) --------2
where
v1=rel velocity of m with respect to M
total energy = constant=
0.5(Mv2square + m(v1-v2)square +k(x1+x2)square) [using 1 and 2]
constant=Mv2square + Msquare*v2square/m +kx2square*(1+M/m)square
=Mv2square(1+M/m) +kx2square(1+M/m)sqaure
=Mv2square +kx2(1+M/m)
taking derivative with respect to time----
2*M*v2*a2=-k(1+M/m)*2*x2*v2
Ma2=-k(m+M)/m*x2
therefore
as a2=-wsquare x2
wsquare=k(M+m)/mM
T=2pie/w
=2pie underroot(Mm/k(m+M))
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