  Question

# A rod of mass m and length l hinged at one end is connected by two springs of spring constants k1 and k2 so that it is horizontal at equilibrium. The angular frequency of the system is (in rad/s)     [Take l=1 m,b=12√2 m,k1=16 N/m,k2=1 N/m,m=164 kg] Solution

## At equilibrium, let elongation and compression of spring 1 and 2 be x1 and x2 respectively. Then, k1x1bcosθ+k2x2lcosθ=mgl2cosθ…(i) [balancing torques about ′O′] Let us rotate the rod by an angle θ. Then, spring 1 gets elongated further by bθ and spring 2 is compressed further by lθ. Writing torque about point ′O′ again, τ0=−[k1bcosθ(x1+bθ)+k2lcosθ(x2+lθ)]+mgl2cosθ  [taking clockwise positive] Using (i), restoring torque about point ′O′ becomes τ0=−(k1bθ×bcosθ+k2lθ×lcosθ) From this, we can say that τ=−(k1b2cosθ+k2l2cosθ)θ But we know that, τ=Iα Moment of inertia of rod about point ′O′ is I=ml23 When θ is very small, cosθ≈1 From this, we can say that, ml23α=−(k1b2+k2l2)θ ⇒α=−3(k1b2+k2l2)ml2θ Comparing this with α=−ω2θ, we get ω=√3(k1b2+k2l2)ml2 From the data given in the question,   ω=      ⎷3×[16×(12√2)2+1×(1)2]164×12 rad/s ⇒ω=24 rad/s  Suggest corrections   