Question

Problem: A simple harmonic oscillator has a block of mass 1.7 kg attached to a spring (k = 260 N/m). When t = 2 s, the block's position and velocity are x = 0.135 m and v = 4.06 m/s. (a) What is the amplitude? What were the (b) position and (c) velocity of the block at zero seconds?

part a = .3549
part b = .2529 m

Please solve for part c.

Answer 1

me 1.7 kg K= 260 Nm 20 = 0.135m V = 4:06 ms at t=2sec Recoll W= k 260 107 = 12.37 Š Now 29- w j az x² 4:06 - 12:37 J AZ_4135² A² = 0.125949 介 So A=0.3549 m Let a: A sincwot + o) T 2 = 0.3549 Sin (12.37+ + OJ As per questim t=2 sec 2 - .135 Sin (12.37 + +0) = 个 So 0135 - 3549 12.37 t +0 = 0.39022 at t=2sec TT o 0.39022 - 12.3742 = 24.35 so 2 = 0.3549 Sin (12.37- 24.35) So å = v= 0.3549 12-37 Cos (12.377 - 24.35) so ht= 2 sec) = 0.3549 *12:37 x cos(-24.35) = 3.1125 mis cans,