A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant 100 N m⁻¹. A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the frequency and the amplitude of the motion.

Question

G1 Physics · Accepted Answer

Initially ω=√(100/1)=10 rad s⁻¹ and at the mean position v=ωA=1 m s⁻¹. When 3 kg is gently placed, horizontal momentum gives v′=(1×1)/(1+3)=0.25 m s⁻¹. New ω′=√(100/4)=5 rad s⁻¹, so f=5/(2π) Hz and A′=v′/ω′=0.05 m=5 cm. All four statements are correct.