Question

Question 2 A particle of mass m, is attached to a spring of natural length 2le and stiffness 2k, and a second spring of stiffness k and natural length lo. It lies on a smooth horizontal table. and the two spring ends are a fixed distance 4lo apart, as shown in Figure Q2. The particle is released from rest at a distance 2lo from each end. Let x be the distance of the particle from A wwwwwwwwww Figure 02

Answer 1

Answer :- 4laner momo min e Answer for "q, - At equilibrium het force on Object 20 elangation in 2k' Spring = (x2 – 200) Here 260 = Natural length of spring elangation in k spring = (ulo-Rombo) Here, lo = Natural length of spring 2k (20 0 1360-Reg) k =) 2K (Keq=200) =(310-8e9K 2 Xeq +X eq = 480t360 xeq = 70

© Determine the total energy E of the mass-spring 2 System. Ang Total energ '' of the mass - Spring System r when it is released from mid point) there were no any kinetic energy. - Total energy = numberof compression in first string + number of Compressions 10 21dstring = { (2K)(0) + { K (210-10) - 4 mu?+K (2-210) +ų kl360-032 Total Energy = 1 klo 1 © Find the speed of from A, in terms of the particle at a distance ✓ lo mik and lo Ang . This is At mean position potential Energy - ļ (2k)(Req=200)*+_ K146-deg-lo)? EK (3lo-2005 + Ke3lo-Il.)? P. E = (0.11% +0.222k)lo P. F = 0.333 klor.

(potential Energy = ☆ klo7 We using. energy conservation Total Energy = kinetic Energy + potential Energy ļ klo a ₂ mv + z klo as I MV2 = klo-ž klo 1 / 2 my2 = 3klo - 2kl2 + v 2 km.” V? 2x klom v² = klo 3m ✓ klo 3m 3 m