Two identical blocks, each of mass m , are connected to two separate massless springs of spring constants k and 3k . Each block is pulled by applying the same constant force F and then released from rest to perform simple harmonic motion. The ratio of the maximum velocity of the block connected to the spring of constant k to that of the block connected to the spring of constant 3k is

Step-by-step solution

When a spring of constant k is stretched by a constant force F , the maximum extension in equilibrium is x = F k . Since the blocks are released from this stretched position, the amplitude of oscillation is A = F k . The maximum velocity of a block in simple harmonic motion is v_ = A . Substituting A = F k and = k m : v_ = ( F k ) k m = F km For the first system with spring constant k₁ = k : v₁ = F km For the second system with spring constant k₂ = 3k : v₂ = F 3km Taking the ratio of their maximum velocities: v₁ v₂