Q 1. A horizontal spring-mass system has a mass of 200 g and spring stiffness of 4.44 N/m. If the mass is given an initial velocity of 0.1 m/s to the right from a starting point 0.05 m to the left of the equilibrium point: 1.1. Give the equation of motion of the mass and the general solution (x(t)) (2 pts) 1.2. Find the natural frequency of the system (1 pts) 1.3. Give the position of the mass at t = 1s (1.5 pts)
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1. The equation of motion for the mass is: x(t2 pts) = 200gMx(0) + 400gMx(1) + 400gMx(2) Show more…
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