Question

A block with mass 0.5 kg is connected by a spring of relaxed length 0.10 m to a post at the center of a low-friction table. You pull the block straight away from the post and release it, and you observe that the period of oscillation is 0.9 s. Next you stretch the spring to a length of 0.22 m and give the block an initial speed v perpendicular to the spring, choosing v so that the motion is a circle with the post at the center. What is this speed v?

Answer 1

Solution :- Natural Length Giren l = 0.1m ma 0.5leg T= oogsec Ammiani system in horizontal direction for a spring-blove Time period & ai Ta ar SE 0 9 - 2* 3.14 / V K= (2x3.14) ²x0.5 (oog)² k= 24.3 N/m NoL Force towards center = KX0.12 = 24.3 20.12 = 30 This force will maintain urcilar motion Om 012m 0,22m1 I= nuo [RE 0.22 m) 3 = 6*5*) ve=1340122 1 3 = R 0.5 xv2 0.22 - 3x0.22 0.5 V v= 1.15 mst