1tmmrw evening will cm tmmrw..........[3][3][3]
Find angular frequency of motion of block of mass m if it is displaced slightly along horz.Neglect inertia effect of rod .Spring constants are k1 and k2.Neglect friction
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1:P abhisekh...
xam tak kya yeh bhi nahi kahoge ?
:P -> distraction thodi na hai ?? ---
:P for power..
:P- for positive...
:P for progress...
:P for perfection...
1bend 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.
IB d2θ/dt2 = F'b + F(a+b) = 0
IB= 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.
1is the ans :
√k1k2/[m{k2(a+b/b)2+k1}] ? [very scary ans]
if yes, then i wil post soln.
or else [2].
1i 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
33[3]
No we will wait... for ur soln... its good manners... so tommorow paste that...[3]
1hey 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 :))
24sooryyyy.......u r correct.....meri aankhen kamzor ho gaye hain...
post the soln/....plz
1[2]
a 'bit' ... woh kitna hota hai ? [3]
ans bata na...
may be did some calculation mistake or typo...