Question

7. -11 points SerCP11 13.P032 My Notes Ask Your Teacher A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 8.40 N is applied. A 0.440-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x = 5.00 cm and released from rest at t = 0, (Assume that the direction of the initial displacement is positive. Use the exact values you enter to make later calculations.) (a) What is the force constant of the spring? N/m (b) What are the angular frequency , the frequency, and the period of the motion? rad/s Hz (c) What is the total energy of the system? (d) What is the amplitude of the motion? cm (e) What are the maximum velocity and the maximum acceleration of the particle? m/s m/s (f) Determine the displacement x of the particle from the equilibrium position att 0.500 s. cm (g) Determine the velocity and acceleration of the particle when t 0.500 s. (Indicate the direction with the sign of your answer.) VE m/s m/s2

I need all of these answers please!!!!!! Will give good rating.

Answer 1

Force = 8.40 x = 3.00 cm = 3 times 10^-2 m mass = m = 0.440 kg Amplitude (A) = 5.00 cm = 5 times 10^-2 m Now F vector = -Kx vector Taking magnitude F = kx Therefore k = F/x = 8.40/3 times 10^-2 = 280 N/m w = squareroot K/m = squareroot 280/0.44 = squareroot 636.36 = 25.23 rad/sec f = w/2 pi = 25.23/2 pi = 4.02 sec^-1 T = pi/w = 1/F = 1/4.02 = 0.249 sec implise Total energy = 1/2 KA^2 = 1/2(280)(5 times 10^-2)^2 = 0.35 Joule implise Amplitude = 5cm implise V_max = wA = (25.23)(5 times 10^-2) = 1.26 m/s a_max = w^2A = (25.23)^@(5 times 10^-2) = 31.83 m/s^@ x = A sin wt t = 0.5 sec Therefore x = A sin[w(0.5)] x = (5cm) sin [25.23(0.5)] x = (5cm) sin[12.615] = (5cm)(0.0486) = 0.243 V = wA cos wt a = -w^2A sin wt t = 0.5 sec Therefore V(0.5) = (25.23)(5 times 10^-2) cos (25.23(0.5)] = 1.26[cos(12.615)] = 1.26(0.9988) = 1.28 m/s [towards positive X] a(0.5) - (2.5.23)^2 (5 times 10^-2) sin[25.23(0.5)] = -31.83[sin[12.615]] = -31.83(0.0486) = -1.547 m/S^2 [towards negative X]