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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your answer is the last one

step-by-step explanation:

divide the length of a'c' by the length of ac to get a scale factor of two

& then reflect vertically (over the y-axis)

so your answer is dilation by a scale factor of 2 followed by reflection over the y-axis

answer:  

step-by-step explanation:

here x represents the height of the pole,

and, the height of the shadow of the pole is 12 feet.

also, the angle of elevation from the top of the pole is 40°,

therefore,

by the low of sine,

which is the required equation for finding the value of x,