Question

A body of mass 1 kg is suspended from a weightless spring having force constant 600 Nm^-1. Another body of mass 0.5 kg moves vertically upwards lift the suspended body with a velocity of 3 ms^-1 and tets embedded in it. Find the frequency of oscillation and amplitude of motion.

Answer 1

Total mass = (1 + 0.5)kg = 1.

k = 600 nm^-1

frequency of oscillations

Let V₁ be velocity of mass after collision

⇒ m₁ V₂ = (m₁ + m₂,) V₁

⇒ 0.5 × 3 = 1.5 V₁

⇒ V₁ = 1 m/s

According to law of conservation of energy

(PE)_max = (KE)_max