Calculate the natural angular frequency ω, in rads-1, of the system shown in Figure. The mass and friction of the pulleys are negligible. Given m=2.5kg,k=20Nm−1

In equilibrium, let the spring is stretched by x0, then T = mg   ……………(1)Also, T+mgsin⁡θ=kx0  ……..(2)⇒ mg+mgsin⁡θ=kx0  ………..(3)Let the spring is further stretched by x then total mechanical energy of the system in this position is,E=12(m+m)v2+12kx+x02−mg(x+xsin⁡θ)Since, E = constant⇒ dEdt=0⇒ 0=2mvdvdt+kx+x0dxdt−mg(1+sin⁡θ)dxdtSubstituting, dxdt=v,dvdt=a and kx0=mg(1+sin⁡θ) we get,a=−k2mxComparing with a=−ω2x, we get          ω=k2m⇒ ω=202(2.5)=205=2rads−1