A body of mass 200g is tied to a spring of spring constant 12.5 N/m, while the other end of spring is fixed at point O. If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed 5 rad/s, then the ratio of extension in the spring to its natural length will be :
Options

Question

A body of mass 200g is tied to a spring of spring constant 12.5 N/m, while the other end of spring is fixed at point O. If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed 5 rad/s, then the ratio of extension in the spring to its natural length will be :

Options

G1 Physics · Accepted Answer

Let natural length be L₀ and extension be x. The radius of circular motion is L₀ + x. The spring force provides centripetal force: kx = m(L₀ + x)ω². Substituting k = 12.5 N/m, m = 0.2 kg and ω = 5 rad/s gives 12.5x = 5(L₀ + x), so 7.5x = 5L₀ and x/L₀ = 2/3.