Question

At equilibrium, the height r of a 0.1 kg mass m hanging on a spring, with negligible mass and spring constant k, is at the position 0. The mass is then set oscillating so that its height x as a function of time is given by x(t)-A . sin(wt + ф-0.1 m . sin(5/s-t + 0.1) Q9. What is the force constant k? (A) 8.2 N/nm (B) 9.1N/nm (C) 2.5 N/m (D) 7.7 N/nm (E) 1.3 N/nm Q10. What is the first time >0 such that 0.05 m? (A) 85 ms (B) 12ms (C) 67 ms (D) 98 ms (E) 50 ms Q11. What is the acceleration at t = 0? (A) -0.25m/s2 (B) -0.45 m/s2 (C) +0.39 m/s2 (D) +0.12 m/s2 (E) +0.25 m/s2

Answer 1

X(t) = A sin (wt + 0) = 0.1 sin(5 st + 0.1) Comparing we get w = 5 rad we have w = squareroot k/m k = w^2 m = (5)^2 times 0.1 k = 2.5 N/m (C) x = 0.05 0.05 = 0.1 sin(5/s t + 0.1) 0.5 = sin(5/s t + 0.1) 5/s t + 0.1 = sin (0.5) = 0.523 t = 0.523 - 0.1/5/s = 0.085 sec t = 85 ms x(t) = 0.1 sin(5/s t + 0.1) a = acceleration = x'' (t) = -(5)^2(0.1) sin(0.5 t + 0.1) at t = 0 a = (5)^2 times 0.1 times sin(0.1) a = -0.25 m/s^2 acceleration = x'' (t) = -w^2 A sin (5/s t + 0.1) Amplitude = w^2 A w^2 A > 9.8 A > 9.8/w^2 = 9.8/(5)^2 = 20.392 A > 39 cm