Kayamat,Nishant sir please help me,,,♥♥

let the extension in the right arm be x₁ and in the left arm be x₂
since the Tension is same every where in the string(as it is massless) we have
3kx_1=kx_2\\ 3x_1=x_2\\ \\ \text{now the total extension is equally shared in both arms,so...} \\ \\ \frac{x_1+x_2}{2}=2x_1\\ \text{this would have produced the same effect if there were a spring of spring constant }\\2k \because\boxed{ 3kx_1+x_1=mg \ \Leftrightarrow \left ( 2k \right )\left ( 2x_1 \right )=mg}\\ \text{So the time period of oscillation is }\boxed{T=2\pi\sqrt{\frac{m}{2k}}}}

P.S --> sorry to answer it as it was addressed to nishant sir or kaymant sir , so the chances of the solution being wrong is high :P

thanks for the try.But the answer you get is not correct.This problem is difficult due to the presence of pulley.If the pulley is not present the problem becomes very easy.There are many problems in H.C. Verma without pulley.Can you have a source to call KAYAMAT or NISHANT sir??If yes please please call them to discuss this problem.Thanks in advance.

so is my answer right ??

it shud be 2pi √m/4k

the answer is 2∩√m/3k

can anyone confirm?

2PI ROOT M/4K

let the displacement of m in vertical direction be x

so the total elongation in both the springs will be 2x

let the elongation in 3k spring be p and in the other one be q (p+q= 2x.....1)

now if the mass m has to stay horizontal the string must slip over the pulley ( i took a lot time to realize the fact)

equating torques 3kp = kq

or 3p= q.....2

so net force = mg+ 3kx (solving 1,2)

so a=-(3k/m) x

so T=2∩√m/3k

a small edit: in post #10 net force means net force given by springs...

@ amrit

as the two springs are symmetrically connected after equating torques

we get 3kp= kq

cancelling k from both sides i wrote 3p=k (As k≠0 :P :P )