Question

A 500-g block is attached to the free end of a spring that has a 5.0-N/m stiffness, while the other end is fixed. It is pulled 15 cm and released. We start recording at t = 0, when it is 8.0 cm to the left of the equilibrium position and moving to the left. (a) What is the phase constant? (b) What is the position, velocity and acceleration of the block immediately after a 2.0-s interval has elapsed? (13 points) ттт Arial 3 (121) T.EE {

Answer 1

KX Since spring will exert a force - kx on the block so md²x dt² d²x + kaca at d3x tw2x=0 dtz which givens x= A Sin (wtto) where de Phase constant. at to, sc = 8 b. A sind - we need to calculate melarity at this instant Since total extension= 15cm. energy & kx²1x5X(0.15)² 0.05625 J. PELAKxf8X102)2 = 0.016 I So fue K.E. = (0.0 605625-0.016) J = 0.04025 = 0.04 025 So tatel energy - at n = 8 2 للفصل of yo mis diffe w.tct times Aw cost wtto) AAS * Aros (6) = 0.13 0.403 05

Squar and add @ and! 2 A2: _82+(01123) A 2.297 A = 2.001 so from Sinda 8 o ggg 874 8.001 89.09 18.col Sin ( ut t 89.09) 7 sor 20 So if a we take wave phase constant, p= 89.09 Sine losing, =(30-89009) 8.91 for o after a dhe t=2 & 2= 8.00 Sin 3.) 62 12 +89.09) = 8.001 Sin (7.879) a7 998 cm. بيا velocity 6.oo low los (wt+89.09) lo > V(2) = 8:0018 uTo cos (auto +6g.og) 25-30 cos (7 879) -0.632 m/s So position= 7. ggf im and velocity 01632 m/s : so after a see,