julielebo8
07.03.2020 •
Mathematics
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.
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Ответ:
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
From newton's second law, we can have the expression for this system as;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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Ответ:
I think it 25/36.
Step-by-step explanation:
The total amount of flour is 4/9 + 1/4. The common denominator of 9 and 4 is 36.
So 4/9 + 1/4 = 16/36 + 9/36 = 25/36