A force of 720 newtons stretches a spring 4 meters. A mass of 45 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 8 m/s. Find the equation of motion.

The final equation of motion of the mass and spring system given in the question is;

x(t) = -4sin 2t

We are given;

Force; F = 720N

displacement; s = 4m

mass; m = 45kg

velocity; v = 8m/s

d²x/dt² + (k/m)x = 0

From hooke's law, we know that;

F = ks

where k is spring constant

k = F/s

k = 720/4

k = 180 N/m

Thus;

d²x/dt² + (720/180)x = 0

d²x/dt² + 4x = 0

The general form of this is;

x'' + ω²x = 0

Thus;

ω² = 4

ω = 2

We know that the solution to it is given by the general expression;

x(t) = c₁ cos ωt + c₂ sinωt

Thus;

x(t) = c₁ cos 2t + c₂ sin2t

When we solve at the initial conditions of the equation, we arrive at;

x(0) = 0 and x'(0) = -8 m/s

Thus, the equation is; x(t) = -4sin 2t

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gimmee one of those gold crowns!

~coco