A 1 kg block executes simple harmonic motion with an amplitude of 0.1 m on a smooth horizontal surface, driven by the restoring force of a spring having a spring constant of 100 N/m . At the instant the block passes through its mean position, a 3 kg block is gently placed on top of it. Assuming the two blocks subsequently move together as a single unit, determine the new frequency and the amplitude of the motion.

Step-by-step solution

Initial mass m₁ = 1 kg , amplitude A₁ = 0.1 m , and spring constant k = 100 N/m . The angular frequency of the initial block is: ₁ = k m₁ = 100 1 = 10 rad/s The velocity of the block at the mean position is: v₁ = A₁ ₁ = 0.1 10 = 1 m/s When the 3 kg block is gently placed on the 1 kg block at the mean position, the horizontal momentum of the system is conserved. Let v₂ be the new velocity at the mean position: m₁ v₁ = (m₁ + m₂) v₂ 1 1 = (1 + 3) v₂ v₂ = 0.25 m/s The new angular frequency of the system is: ₂ = k m₁ +