SHM......gud

tmmrw evening will cm tmmrw..........[3][3][3]

:O

bend the rod a litttttttle.

let, x1=extension in spring1

x2= extension in spring 2

x = x1 + x2

x2=F/k2

x1=F'/k1 . (a+b/ b)

now, torque about B.

I_B d²Î¸/dt² = F'b + F(a+b) = 0

I_B= 0 (given in question .... to neglect..)

so, |F'| = F(a+b)/b

so, x1= F(a+b)²/k1.b²

x=x1+x2 = - [ F(a+b/b)²/k1 + F/k2 ]

now write in F=-Kx form..

where F=mω²x.

hey sky...............plzzzzz post the answer dear..............

is the ans :

√k1k2/[m{k2(a+b/b)²+k1}] ? [very scary ans]

if yes, then i wil post soln.

or else [2].

i am keen to answer this but have a silly doubt can u pls explain me hows it moved i mean horizontally but how? at pt where b is there ? is it? pls explain

[3]

No we will wait... for ur soln... its good manners... so tommorow paste that...[3]

hey will u pls wait till tomorrow evening.. ??

in the mean tym if someone posts... i will be tooo happy...

see, its a long one..

i am too embarrassed for such a thing... but plz.. (kal na exam hai actually :))

[2]

a 'bit' ... woh kitna hota hai ? [3]

ans bata na...

may be did some calculation mistake or typo...

a bit wrong.......