SHM at its best!!!!!

First i will rite an energy equation for a small oscillation for the whole system and then diff it to get the 'w'

this is the best question ive ever found in SHM .... and the sense of satisfaction u get after solvin it is immense!!!!![1][1][1][1][1][1][1]

doing it by force method is way easier here sankara......!!!!!

arey main yahii likh rahi thi abhi abhi ...

kk :)

if noone attempts [2] solutions will be posted by tommorow !!

ye A M² kya hai?????????

exactly wo solid ke barre mein kya lika hai

kahi ye density to nahi hai

ok....i think its area...........A metre²[12]

we'll ablnce d torues let d small displcmnt b x so let d tension b T so TR-kxR/2=Ialpha and T+l^2xρ=mg yeh dohno solve karke woh x ke terms emin gadbad ho ri hai :(

here i assumed l as lenght of d cube

its the distance of the spring from the center of the cylinder .... A is the area of crossection........ get ans. in terms of watever is given only!!!!!

HINT its only ΔT that matters !!!!!!

ok......iit..........give some more time..........i couldnt understand the fig..........now i will try.....[1][1]

heyy cmon sum 1 try!!!!!![2] ..... okie ill post soln.

okie now as i gave in my hint we will concider ΔT only!!!!!!!!!!

now let us take the block to be pushed in by distance x................

an extra ρgAx acts upward and also theres an accn. say a upward !!!!!!!!!!

so ΔT + ρgAx =ma.........

ΔT = ma - ρgAx ........

x = Rθ if the cylinder is displaced by an angle θ

................

now comin to the spring .........

if the disp. of the cylinder is θ ......

the extension in spring is R/2θ .......

so Fsp = kR/2θ .............

now coming to torques acting on the cylinder......

kR/2θ(R/2) - ΔTR = (MR²/2)(ω²Î¸)

kRθ/4 - ma + ρgARθ = (mR/2)(ω²Î¸)

kRθ/4 - mω²(Rθ) + ρgARθ = mR/2(ω²Î¸)

k/4 + ρgA = 3/2 m(ω² )......

k+4ρgA /6m = ω² ...

√k+4ρgA /6m = ω .............

if u guys hav simpler methods do post!!!!![1]